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Notes

Questions

Find all numbers c that satisfy the conclusions of the mean value theorem.

\(\textbf{1)}\) \( f(x)=x^2 \, \) on \( \, [-3,2] \)

Show Answer

The answer is \( c=-1/2 \)

Show Work

\(\text{Step 1: Notes}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{f(b)-f(a)}{b-a}\)


\(\text{Step 2: Plug in }a=-3 \text{ and }b=2\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{f(2)-f(-3)}{(2)-(-3)}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{(2)^2-(-3)^2}{2+3}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{4-9}{5}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{-5}{5}\)

\(\,\,\,\,\,\,f'(c)=-1\)


\(\text{Step 3: Find }f'(c)\)

\(\,\,\,\,\,\,f(x)=x^2\)

\(\,\,\,\,\,\,f'(x)=2x\)

\(\,\,\,\,\,\,f'(c)=2c\)


\(\text{Step 4: Plug in }f'(c)=2c\)

\(\,\,\,\,\,\,f'(c)=-1\)

\(\,\,\,\,\,\,2c=-1\)


\(\text{Step 5: Solve for }c\)

\(\,\,\,\,\,\,2c=-1\)

\(\,\,\,\,\,\,c=\,-\,\frac{1}{2}\)

\(\,\,\,\,\,\,\)The answer is \( c=-1/2 \)

\(\textbf{2)}\) \( f(x)=x^3+2x+4 \,\) on \( \, [0,2] \)

Show Answer

The answer is \( c=\displaystyle\frac{2\sqrt{3}}{3} \)

Show Work

\(\text{Step 1: Notes}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{f(b)-f(a)}{b-a}\)


\(\text{Step 2: Plug in }a=0 \text{ and }b=2\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{f(2)-f(0)}{(2)-(0)}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{(2)^3+2(2)+4-(0)^3-2(0)-4}{2-0}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{8+4}{2}\)

\(\,\,\,\,\,\,f'(c)=\displaystyle \frac{12}{2}\)

\(\,\,\,\,\,\,f'(c)=6\)


\(\text{Step 3: Find }f'(c)\)

\(\,\,\,\,\,\,f(x)=x^3+2x+4\)

\(\,\,\,\,\,\,f'(x)=3x^2+2\)

\(\,\,\,\,\,\,f'(c)=3c^2+2\)


\(\text{Step 4: Plug in }f'(c)=3c^2+2\)

\(\,\,\,\,\,\,f'(c)=6\)

\(\,\,\,\,\,\,3c^2+2=6\)


\(\text{Step 5: Solve for }c\)

\(\,\,\,\,\,\,3c^2+2=6\)

\(\,\,\,\,\,\,3c^2=6-2\)

\(\,\,\,\,\,\,3c^2=4\)

\(\,\,\,\,\,\,c^2=\displaystyle\frac{4}{3}\)

\(\,\,\,\,\,\,c=\pm\sqrt{\displaystyle\frac{4}{3}}\)

\(\,\,\,\,\,\,c=\pm\displaystyle\frac{2\sqrt{3}}{3}\)

\(\,\,\,\,\,\,\)The answer is \( c=\displaystyle\frac{2\sqrt{3}}{3} \)

\(\textbf{3)}\) \( f(x)=\sin⁡ x \,\) on \( \, [0,\pi] \)

Show Answer

The answer is \( c=\displaystyle\frac{\pi}{2} \)

\(\textbf{4)}\) \( f(x)=x^2 \,\) on \( \, [-1,1] \)

Show Answer

The answer is \( c=0 \)

See Related Pages\(\)

\(\bullet\text{Mean Value Theorem Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(emathhelp.net)}\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)

\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)

\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)

\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)

\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)

\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)

\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)

\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)

\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)

\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)

\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)

\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)

In Summary

The mean value theorem for derivatives states that if a function f is continuous and differentiable on the interval [a, b], then there exists at least one point c in the interval (a, b) such that the derivative of f at x=c is equal to the average rate of change of f on the interval [a, b].

The derivative represents the instantaneous slope of a function at a given point. So this theorem is stating that as long as the function is continuous and differentiable on the entire interval between the endpoints there must be at least one point where the
instantaneous slope is the same as the slope through the endpoints of the interval.

The mean value theorem is used to prove other results in calculus, such as the intermediate value theorem and the extreme value theorem.

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