Percentage rate of change compares the instantaneous rate of change of a function to the size of the function itself. In calculus, it is found by dividing the derivative by the original function value, then converting the decimal to a percent. These problems focus on finding \(\frac{f'(x)}{f(x)}\) at a specific input and interpreting the result as a percentage.
Notes
Percentage Rate of Change
\(\displaystyle\frac{f'(x)}{f(x)}\)
Practice Problems
\(\textbf{1)}\) \(\text{Find the percentage rate of change of } f(x)=4x^2+200 \text{ at } x=2.\)
\(\text{The answer is } \approx 7.41\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=8x\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{8x}{4x^2+200}\)
\(\,\,\,\,\,\,\text{Plug in }x=2:\displaystyle \frac{8(2)}{4(2)^2+200}\)
\(\,\,\,\,\,\,\displaystyle \frac{16}{216}\approx 0.074074\)
\(\,\,\,\,\,\,0.074074\cdot100\%\approx 7.41\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx 7.41\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=8x\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{8x}{4x^2+200}\)
\(\,\,\,\,\,\,\text{Plug in }x=2:\displaystyle \frac{8(2)}{4(2)^2+200}\)
\(\,\,\,\,\,\,\displaystyle \frac{16}{216}\approx 0.074074\)
\(\,\,\,\,\,\,0.074074\cdot100\%\approx 7.41\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx 7.41\% \)
\(\textbf{2)}\) \(\text{Find the percentage rate of change of } f(x)=3x+20 \text{ at } x=3.\)
\(\text{The answer is } \approx 10.34\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=3\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{3}{3x+20}\)
\(\,\,\,\,\,\,\text{Plug in }x=3:\displaystyle \frac{3}{3(3)+20}\)
\(\,\,\,\,\,\,\displaystyle \frac{3}{29}\approx 0.1034\)
\(\,\,\,\,\,\,0.1034\cdot100\%\approx 10.34\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx 10.34\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=3\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{3}{3x+20}\)
\(\,\,\,\,\,\,\text{Plug in }x=3:\displaystyle \frac{3}{3(3)+20}\)
\(\,\,\,\,\,\,\displaystyle \frac{3}{29}\approx 0.1034\)
\(\,\,\,\,\,\,0.1034\cdot100\%\approx 10.34\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx 10.34\% \)
\(\textbf{3)}\) \(\text{Find the percentage rate of change of } f(x)=2x^3+4x+1 \text{ at } x=1.\)
\(\text{The answer is } \approx 142.9\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=6x^2+4\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{6x^2+4}{2x^3+4x+1}\)
\(\,\,\,\,\,\,\text{Plug in }x=1:\displaystyle \frac{6(1)^2+4}{2(1)^3+4(1)+1}\)
\(\,\,\,\,\,\,\displaystyle \frac{10}{7}\approx 1.429\)
\(\,\,\,\,\,\,1.429\cdot100\%\approx 142.9\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx 142.9\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=6x^2+4\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{6x^2+4}{2x^3+4x+1}\)
\(\,\,\,\,\,\,\text{Plug in }x=1:\displaystyle \frac{6(1)^2+4}{2(1)^3+4(1)+1}\)
\(\,\,\,\,\,\,\displaystyle \frac{10}{7}\approx 1.429\)
\(\,\,\,\,\,\,1.429\cdot100\%\approx 142.9\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx 142.9\% \)
\(\textbf{4)}\) \(\text{Find the percentage rate of change of } f(x)=5x^2+50 \text{ at } x=5.\)
\(\text{The answer is } \approx 28.57\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=10x\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{10x}{5x^2+50}\)
\(\,\,\,\,\,\,\text{Plug in }x=5:\displaystyle \frac{10(5)}{5(5)^2+50}\)
\(\,\,\,\,\,\,\displaystyle \frac{50}{175}\approx0.2857\)
\(\,\,\,\,\,\,0.2857\cdot100\%\approx28.57\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx28.57\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=10x\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{10x}{5x^2+50}\)
\(\,\,\,\,\,\,\text{Plug in }x=5:\displaystyle \frac{10(5)}{5(5)^2+50}\)
\(\,\,\,\,\,\,\displaystyle \frac{50}{175}\approx0.2857\)
\(\,\,\,\,\,\,0.2857\cdot100\%\approx28.57\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx28.57\%\)
\(\textbf{5)}\) \(\text{Find the percentage rate of change of } f(x)=100e^{0.04x} \text{ at } x=10.\)
\(\text{The answer is }4\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=4e^{0.04x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{4e^{0.04x}}{100e^{0.04x}}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=0.04\)
\(\,\,\,\,\,\,0.04\cdot100\%=4\%\)
\(\,\,\,\,\,\,\)The answer is \(4\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=4e^{0.04x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{4e^{0.04x}}{100e^{0.04x}}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=0.04\)
\(\,\,\,\,\,\,0.04\cdot100\%=4\%\)
\(\,\,\,\,\,\,\)The answer is \(4\%\)
\(\textbf{6)}\) \(\text{Find the percentage rate of change of } f(x)=200(1.03)^x \text{ at } x=8.\)
\(\text{The answer is } \approx 2.96\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=200(1.03)^x\ln(1.03)\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{200(1.03)^x\ln(1.03)}{200(1.03)^x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\ln(1.03)\)
\(\,\,\,\,\,\,\ln(1.03)\approx0.02956\)
\(\,\,\,\,\,\,0.02956\cdot100\%\approx2.96\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx2.96\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=200(1.03)^x\ln(1.03)\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{200(1.03)^x\ln(1.03)}{200(1.03)^x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\ln(1.03)\)
\(\,\,\,\,\,\,\ln(1.03)\approx0.02956\)
\(\,\,\,\,\,\,0.02956\cdot100\%\approx2.96\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx2.96\%\)
\(\textbf{7)}\) \(\text{Find the percentage rate of change of } f(x)=\ln{x} \text{ at } x=e.\)
\(\text{The answer is } \approx 36.79\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{\frac{1}{x}}{\ln{x}}\)
\(\,\,\,\,\,\,\text{Plug in }x=e:\displaystyle \frac{\frac{1}{e}}{\ln{e}}\)
\(\,\,\,\,\,\,\displaystyle \frac{1/e}{1}=\frac{1}{e}\approx0.3679\)
\(\,\,\,\,\,\,0.3679\cdot100\%\approx36.79\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx36.79\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{\frac{1}{x}}{\ln{x}}\)
\(\,\,\,\,\,\,\text{Plug in }x=e:\displaystyle \frac{\frac{1}{e}}{\ln{e}}\)
\(\,\,\,\,\,\,\displaystyle \frac{1/e}{1}=\frac{1}{e}\approx0.3679\)
\(\,\,\,\,\,\,0.3679\cdot100\%\approx36.79\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx36.79\%\)
\(\textbf{8)}\) \(\text{Find the percentage rate of change of } f(x)=x^3+1 \text{ at } x=2.\)
\(\text{The answer is } \approx 133.33\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=3x^2\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{3x^2}{x^3+1}\)
\(\,\,\,\,\,\,\text{Plug in }x=2:\displaystyle \frac{3(2)^2}{2^3+1}\)
\(\,\,\,\,\,\,\displaystyle \frac{12}{9}=\frac{4}{3}\approx1.3333\)
\(\,\,\,\,\,\,1.3333\cdot100\%\approx133.33\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx133.33\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=3x^2\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{3x^2}{x^3+1}\)
\(\,\,\,\,\,\,\text{Plug in }x=2:\displaystyle \frac{3(2)^2}{2^3+1}\)
\(\,\,\,\,\,\,\displaystyle \frac{12}{9}=\frac{4}{3}\approx1.3333\)
\(\,\,\,\,\,\,1.3333\cdot100\%\approx133.33\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx133.33\%\)
\(\textbf{9)}\) \(\text{Find the percentage rate of change of } f(x)=500-20x \text{ at } x=5.\)
\(\text{The answer is }-5\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=-20\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{-20}{500-20x}\)
\(\,\,\,\,\,\,\text{Plug in }x=5:\displaystyle \frac{-20}{500-20(5)}\)
\(\,\,\,\,\,\,\displaystyle \frac{-20}{400}=-0.05\)
\(\,\,\,\,\,\,-0.05\cdot100\%=-5\%\)
\(\,\,\,\,\,\,\)The answer is \(-5\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=-20\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{-20}{500-20x}\)
\(\,\,\,\,\,\,\text{Plug in }x=5:\displaystyle \frac{-20}{500-20(5)}\)
\(\,\,\,\,\,\,\displaystyle \frac{-20}{400}=-0.05\)
\(\,\,\,\,\,\,-0.05\cdot100\%=-5\%\)
\(\,\,\,\,\,\,\)The answer is \(-5\%\)
\(\textbf{10)}\) \(\text{Find the percentage rate of change of } f(x)=\sqrt{x}+10 \text{ at } x=16.\)
\(\text{The answer is } \approx 0.89\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{\frac{1}{2\sqrt{x}}}{\sqrt{x}+10}\)
\(\,\,\,\,\,\,\text{Plug in }x=16:\displaystyle \frac{\frac{1}{2\sqrt{16}}}{\sqrt{16}+10}\)
\(\,\,\,\,\,\,\displaystyle \frac{\frac{1}{8}}{14}=\frac{1}{112}\approx0.00893\)
\(\,\,\,\,\,\,0.00893\cdot100\%\approx0.89\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx0.89\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{\frac{1}{2\sqrt{x}}}{\sqrt{x}+10}\)
\(\,\,\,\,\,\,\text{Plug in }x=16:\displaystyle \frac{\frac{1}{2\sqrt{16}}}{\sqrt{16}+10}\)
\(\,\,\,\,\,\,\displaystyle \frac{\frac{1}{8}}{14}=\frac{1}{112}\approx0.00893\)
\(\,\,\,\,\,\,0.00893\cdot100\%\approx0.89\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx0.89\%\)
\(\textbf{11)}\) \(\text{Find the percentage rate of change of } f(x)=\frac{x}{x+2} \text{ at } x=2.\)
\(\text{The answer is }25\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{(x+2)(1)-x(1)}{(x+2)^2}\)
\(\,\,\,\,\,\,f'(x)=\frac{2}{(x+2)^2}\)
\(\,\,\,\,\,\,f(2)=\frac{2}{4}=\frac{1}{2}\)
\(\,\,\,\,\,\,f'(2)=\frac{2}{4^2}=\frac{1}{8}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(2)}{f(2)}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}=0.25\)
\(\,\,\,\,\,\,0.25\cdot100\%=25\%\)
\(\,\,\,\,\,\,\)The answer is \(25\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{(x+2)(1)-x(1)}{(x+2)^2}\)
\(\,\,\,\,\,\,f'(x)=\frac{2}{(x+2)^2}\)
\(\,\,\,\,\,\,f(2)=\frac{2}{4}=\frac{1}{2}\)
\(\,\,\,\,\,\,f'(2)=\frac{2}{4^2}=\frac{1}{8}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(2)}{f(2)}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}=0.25\)
\(\,\,\,\,\,\,0.25\cdot100\%=25\%\)
\(\,\,\,\,\,\,\)The answer is \(25\%\)
\(\textbf{12)}\) \(\text{Find the percentage rate of change of } f(x)=(x+1)^2 \text{ at } x=4.\)
\(\text{The answer is }40\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=2(x+1)\)
\(\,\,\,\,\,\,f'(4)=2(5)=10\)
\(\,\,\,\,\,\,f(4)=(4+1)^2=25\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(4)}{f(4)}=\frac{10}{25}=0.4\)
\(\,\,\,\,\,\,0.4\cdot100\%=40\%\)
\(\,\,\,\,\,\,\)The answer is \(40\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=2(x+1)\)
\(\,\,\,\,\,\,f'(4)=2(5)=10\)
\(\,\,\,\,\,\,f(4)=(4+1)^2=25\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(4)}{f(4)}=\frac{10}{25}=0.4\)
\(\,\,\,\,\,\,0.4\cdot100\%=40\%\)
\(\,\,\,\,\,\,\)The answer is \(40\%\)
\(\textbf{13)}\) \(\text{Find the percentage rate of change of } f(x)=\frac{50}{x+5} \text{ at } x=5.\)
\(\text{The answer is }-10\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f(x)=50(x+5)^{-1}\)
\(\,\,\,\,\,\,f'(x)=-50(x+5)^{-2}\)
\(\,\,\,\,\,\,f'(5)=-\frac{50}{10^2}=-\frac{1}{2}\)
\(\,\,\,\,\,\,f(5)=\frac{50}{10}=5\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(5)}{f(5)}=\frac{-\frac{1}{2}}{5}=-0.1\)
\(\,\,\,\,\,\,-0.1\cdot100\%=-10\%\)
\(\,\,\,\,\,\,\)The answer is \(-10\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f(x)=50(x+5)^{-1}\)
\(\,\,\,\,\,\,f'(x)=-50(x+5)^{-2}\)
\(\,\,\,\,\,\,f'(5)=-\frac{50}{10^2}=-\frac{1}{2}\)
\(\,\,\,\,\,\,f(5)=\frac{50}{10}=5\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(5)}{f(5)}=\frac{-\frac{1}{2}}{5}=-0.1\)
\(\,\,\,\,\,\,-0.1\cdot100\%=-10\%\)
\(\,\,\,\,\,\,\)The answer is \(-10\%\)
\(\textbf{14)}\) \(\text{Find the percentage rate of change of } f(x)=x^2e^x \text{ at } x=1.\)
\(\text{The answer is }300\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=2xe^x+x^2e^x\)
\(\,\,\,\,\,\,f'(x)=e^x(2x+x^2)\)
\(\,\,\,\,\,\,f'(1)=e(2+1)=3e\)
\(\,\,\,\,\,\,f(1)=1^2e^1=e\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(1)}{f(1)}=\frac{3e}{e}=3\)
\(\,\,\,\,\,\,3\cdot100\%=300\%\)
\(\,\,\,\,\,\,\)The answer is \(300\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=2xe^x+x^2e^x\)
\(\,\,\,\,\,\,f'(x)=e^x(2x+x^2)\)
\(\,\,\,\,\,\,f'(1)=e(2+1)=3e\)
\(\,\,\,\,\,\,f(1)=1^2e^1=e\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(1)}{f(1)}=\frac{3e}{e}=3\)
\(\,\,\,\,\,\,3\cdot100\%=300\%\)
\(\,\,\,\,\,\,\)The answer is \(300\%\)
\(\textbf{15)}\) \(\text{Find the percentage rate of change of } f(x)=e^{-0.2x} \text{ at } x=4.\)
\(\text{The answer is }-20\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=-0.2e^{-0.2x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{-0.2e^{-0.2x}}{e^{-0.2x}}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=-0.2\)
\(\,\,\,\,\,\,-0.2\cdot100\%=-20\%\)
\(\,\,\,\,\,\,\)The answer is \(-20\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=-0.2e^{-0.2x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{-0.2e^{-0.2x}}{e^{-0.2x}}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=-0.2\)
\(\,\,\,\,\,\,-0.2\cdot100\%=-20\%\)
\(\,\,\,\,\,\,\)The answer is \(-20\%\)
\(\textbf{16)}\) \(\text{Find the percentage rate of change of } f(x)=5x^4-3x \text{ at } x=1.\)
\(\text{The answer is }850\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=20x^3-3\)
\(\,\,\,\,\,\,f'(1)=20(1)^3-3=17\)
\(\,\,\,\,\,\,f(1)=5(1)^4-3(1)=2\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(1)}{f(1)}=\frac{17}{2}=8.5\)
\(\,\,\,\,\,\,8.5\cdot100\%=850\%\)
\(\,\,\,\,\,\,\)The answer is \(850\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=20x^3-3\)
\(\,\,\,\,\,\,f'(1)=20(1)^3-3=17\)
\(\,\,\,\,\,\,f(1)=5(1)^4-3(1)=2\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(1)}{f(1)}=\frac{17}{2}=8.5\)
\(\,\,\,\,\,\,8.5\cdot100\%=850\%\)
\(\,\,\,\,\,\,\)The answer is \(850\%\)
\(\textbf{17)}\) \(\text{Find the percentage rate of change of } f(x)=1000(0.98)^x \text{ at } x=6.\)
\(\text{The answer is } \approx -2.02\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=1000(0.98)^x\ln(0.98)\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{1000(0.98)^x\ln(0.98)}{1000(0.98)^x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\ln(0.98)\)
\(\,\,\,\,\,\,\ln(0.98)\approx-0.0202\)
\(\,\,\,\,\,\,-0.0202\cdot100\%\approx-2.02\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx-2.02\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=1000(0.98)^x\ln(0.98)\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\frac{1000(0.98)^x\ln(0.98)}{1000(0.98)^x}\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(x)}{f(x)}=\ln(0.98)\)
\(\,\,\,\,\,\,\ln(0.98)\approx-0.0202\)
\(\,\,\,\,\,\,-0.0202\cdot100\%\approx-2.02\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx-2.02\%\)
\(\textbf{18)}\) \(\text{Find the percentage rate of change of } f(x)=4x^2+12x+9 \text{ at } x=3.\)
\(\text{The answer is } \approx 44.44\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=8x+12\)
\(\,\,\,\,\,\,f'(3)=8(3)+12=36\)
\(\,\,\,\,\,\,f(3)=4(3)^2+12(3)+9\)
\(\,\,\,\,\,\,f(3)=36+36+9=81\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(3)}{f(3)}=\frac{36}{81}\approx0.4444\)
\(\,\,\,\,\,\,0.4444\cdot100\%\approx44.44\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx44.44\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=8x+12\)
\(\,\,\,\,\,\,f'(3)=8(3)+12=36\)
\(\,\,\,\,\,\,f(3)=4(3)^2+12(3)+9\)
\(\,\,\,\,\,\,f(3)=36+36+9=81\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(3)}{f(3)}=\frac{36}{81}\approx0.4444\)
\(\,\,\,\,\,\,0.4444\cdot100\%\approx44.44\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx44.44\%\)
\(\textbf{19)}\) \(\text{Find the percentage rate of change of } f(x)=\sqrt{3x+1} \text{ at } x=5.\)
\(\text{The answer is } \approx 9.38\% \)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{3}{2\sqrt{3x+1}}\)
\(\,\,\,\,\,\,f'(5)=\frac{3}{2\sqrt{16}}=\frac{3}{8}\)
\(\,\,\,\,\,\,f(5)=\sqrt{16}=4\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(5)}{f(5)}=\frac{\frac{3}{8}}{4}=\frac{3}{32}=0.09375\)
\(\,\,\,\,\,\,0.09375\cdot100\%=9.375\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx9.38\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{3}{2\sqrt{3x+1}}\)
\(\,\,\,\,\,\,f'(5)=\frac{3}{2\sqrt{16}}=\frac{3}{8}\)
\(\,\,\,\,\,\,f(5)=\sqrt{16}=4\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(5)}{f(5)}=\frac{\frac{3}{8}}{4}=\frac{3}{32}=0.09375\)
\(\,\,\,\,\,\,0.09375\cdot100\%=9.375\%\)
\(\,\,\,\,\,\,\)The answer is \(\approx9.38\%\)
\(\textbf{20)}\) \(\text{Find the percentage rate of change of } f(x)=\frac{x^2+1}{x+1} \text{ at } x=1.\)
\(\text{The answer is }50\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{(x+1)(2x)-(x^2+1)(1)}{(x+1)^2}\)
\(\,\,\,\,\,\,f'(x)=\frac{2x^2+2x-x^2-1}{(x+1)^2}\)
\(\,\,\,\,\,\,f'(x)=\frac{x^2+2x-1}{(x+1)^2}\)
\(\,\,\,\,\,\,f'(1)=\frac{1+2-1}{(2)^2}=\frac{1}{2}\)
\(\,\,\,\,\,\,f(1)=\frac{1^2+1}{1+1}=1\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(1)}{f(1)}=\frac{1/2}{1}=0.5\)
\(\,\,\,\,\,\,0.5\cdot100\%=50\%\)
\(\,\,\,\,\,\,\)The answer is \(50\%\)
\(\,\,\,\,\,\,\text{Percentage rate of change is }\displaystyle \frac{f'(x)}{f(x)}\)
\(\,\,\,\,\,\,f'(x)=\frac{(x+1)(2x)-(x^2+1)(1)}{(x+1)^2}\)
\(\,\,\,\,\,\,f'(x)=\frac{2x^2+2x-x^2-1}{(x+1)^2}\)
\(\,\,\,\,\,\,f'(x)=\frac{x^2+2x-1}{(x+1)^2}\)
\(\,\,\,\,\,\,f'(1)=\frac{1+2-1}{(2)^2}=\frac{1}{2}\)
\(\,\,\,\,\,\,f(1)=\frac{1^2+1}{1+1}=1\)
\(\,\,\,\,\,\,\displaystyle \frac{f'(1)}{f(1)}=\frac{1/2}{1}=0.5\)
\(\,\,\,\,\,\,0.5\cdot100\%=50\%\)
\(\,\,\,\,\,\,\)The answer is \(50\%\)
See Related Pages\(\)
\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)
\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)
\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)
\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)
\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)
\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)
\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)
\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)
\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)
\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)
\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)