A horizontal tangent line occurs when the slope of a function is zero at a point on the graph. To find horizontal tangent lines, take the derivative, set it equal to zero, solve for the x-values, and then plug those x-values back into the original function. The equations of horizontal tangent lines are always in the form \(y=c\).
Notes

Practice Problems
Find the equations of the horizontal tangent lines.
\(\textbf{1)}\) \( f(x)=x^2+4x+4 \)
\(\,\,\,\,\,f'(x)=2x+4\)
\(\,\,\,\,\,2x+4=0\)
\(\,\,\,\,\,2x=-4\)
\(\,\,\,\,\,x=-2\)
\(\,\,\,\,\,f(-2)=(-2)^2+4(-2)+4\)
\(\,\,\,\,\,f(-2)=4-8+4\)
\(\,\,\,\,\,f(-2)=0\)
\(\,\,\,\,\,\)The answer is \(y=0\)
The answer is \( y=0 \)
\(\textbf{2)}\) \( f(x)=\sin x \)
\(\,\,\,\,\,f'(x)=\cos{x}\)
\(\,\,\,\,\,\cos{x}=0\)
\(\,\,\,\,\,x=\frac{\pi}{2}+n\pi\)
\(\,\,\,\,\,\sin\left(\frac{\pi}{2}\right)=1\)
\(\,\,\,\,\,\sin\left(\frac{3\pi}{2}\right)=-1\)
\(\,\,\,\,\,\)The answers are \(y=1\) and \(y=-1\)
The answer is \( y=1 \) and \( y=-1 \)
\(\textbf{3)}\) \( f(x)=4 \)
\(\,\,\,\,\,f(x)=4\)
\(\,\,\,\,\,f'(x)=0\)
\(\,\,\,\,\,\text{The derivative is }0\text{ for every }x.\)
\(\,\,\,\,\,\text{The entire function is already horizontal.}\)
\(\,\,\,\,\,\)The answer is \(y=4\)
The answer is \( y=4 \)
\(\textbf{4)}\) \( f(x) = x^3 – 3x^2 + 3x \)
\(\,\,\,\,\,f'(x) = 3x^2 – 6x + 3\)
\(\,\,\,\,\,3x^2 – 6x + 3 = 0\)
\(\,\,\,\,\,3(x^2 – 2x + 1) = 0\)
\(\,\,\,\,\,(x – 1)^2 = 0\)
\(\,\,\,\,\,x = 1\)
\(\,\,\,\,\,f(1) = (1)^3 – 3(1)^2 + 3(1)\)
\(\,\,\,\,\,f(1)=1 – 3 + 3 = 1\)
\(\,\,\,\,\,\)The answer is \(y=1\)
The answer is \( y=1 \)
\(\textbf{5)}\) \( f(x) = x^2 – 4 \)
\(\,\,\,\,\,f'(x) = 2x\)
\(\,\,\,\,\,2x = 0\)
\(\,\,\,\,\,x = 0\)
\(\,\,\,\,\,f(0) = (0)^2 – 4\)
\(\,\,\,\,\,f(0)=-4\)
\(\,\,\,\,\,\)The answer is \(y=-4\)
The answer is \( y=-4 \)
\(\textbf{6)}\) \( f(x) = x^3 – 3x \)
\(\,\,\,\,\,f'(x) = 3x^2 – 3\)
\(\,\,\,\,\,3x^2 – 3 = 0\)
\(\,\,\,\,\,3(x^2 – 1) = 0\)
\(\,\,\,\,\,x^2 = 1\)
\(\,\,\,\,\,x = \pm 1\)
\(\,\,\,\,\,f(1) = (1)^3 – 3(1) = -2\)
\(\,\,\,\,\,f(-1) = (-1)^3 – 3(-1) = 2\)
\(\,\,\,\,\,\)The answers are \(y=-2\) and \(y=2\)
The answers are \( y = -2 \) and \( y = 2 \)
\(\textbf{7)}\) \( f(x) = \cos x \)
\(\,\,\,\,\,f'(x) = -\sin x\)
\(\,\,\,\,\, -\sin x = 0 \)
\(\,\,\,\,\, \sin x = 0 \)
\(\,\,\,\,\, x = n\pi \)
\(\,\,\,\,\,f(0) = \cos(0) = 1\)
\(\,\,\,\,\,f(\pi) = \cos(\pi) = -1\)
\(\,\,\,\,\,\)The answers are \(y=1\) and \(y=-1\)
The answers are \( y=1 \) and \( y=-1 \)
\(\textbf{8)}\) \( f(x) = 2x^2 – 8x + 6 \)
\(\,\,\,\,\,f'(x) = 4x – 8\)
\(\,\,\,\,\,4x – 8 = 0\)
\(\,\,\,\,\,4x = 8\)
\(\,\,\,\,\,x = 2\)
\(\,\,\,\,\,f(2) = 2(2)^2 – 8(2) + 6\)
\(\,\,\,\,\,f(2)=8 – 16 + 6 = -2\)
\(\,\,\,\,\,\)The answer is \(y=-2\)
The answer is \( y=-2 \)
\(\textbf{9)}\) \( f(x)=x^3-6x^2+9x \)
\(\,\,\,\,\,f'(x)=3x^2-12x+9\)
\(\,\,\,\,\,3x^2-12x+9=0\)
\(\,\,\,\,\,3(x^2-4x+3)=0\)
\(\,\,\,\,\,3(x-1)(x-3)=0\)
\(\,\,\,\,\,x=1\text{ or }x=3\)
\(\,\,\,\,\,f(1)=1-6+9=4\)
\(\,\,\,\,\,f(3)=27-54+27=0\)
\(\,\,\,\,\,\)The answers are \(y=4\) and \(y=0\)
The answers are \( y=4 \) and \( y=0 \)
\(\textbf{10)}\) \( f(x)=x^4-8x^2 \)
\(\,\,\,\,\,f'(x)=4x^3-16x\)
\(\,\,\,\,\,4x^3-16x=0\)
\(\,\,\,\,\,4x(x^2-4)=0\)
\(\,\,\,\,\,4x(x-2)(x+2)=0\)
\(\,\,\,\,\,x=-2,0,2\)
\(\,\,\,\,\,f(-2)=16-32=-16\)
\(\,\,\,\,\,f(0)=0\)
\(\,\,\,\,\,f(2)=16-32=-16\)
\(\,\,\,\,\,\)The answers are \(y=-16\) and \(y=0\)
The answers are \( y=-16 \) and \( y=0 \)
\(\textbf{11)}\) \( f(x)=e^x \)
\(\,\,\,\,\,f'(x)=e^x\)
\(\,\,\,\,\,e^x=0\)
\(\,\,\,\,\,\text{There is no real value of }x\text{ where }e^x=0.\)
\(\,\,\,\,\,\text{So there are no horizontal tangent lines.}\)
\(\,\,\,\,\,\)There are no horizontal tangent lines.
There are no horizontal tangent lines.
\(\textbf{12)}\) \( f(x)=\ln{x} \)
\(\,\,\,\,\,f'(x)=\frac{1}{x}\)
\(\,\,\,\,\,\frac{1}{x}=0\)
\(\,\,\,\,\,\text{There is no value of }x\text{ where }\frac{1}{x}=0.\)
\(\,\,\,\,\,\text{So there are no horizontal tangent lines.}\)
\(\,\,\,\,\,\)There are no horizontal tangent lines.
There are no horizontal tangent lines.
\(\textbf{13)}\) \( f(x)=x+\frac{4}{x} \)
\(\,\,\,\,\,f'(x)=1-\frac{4}{x^2}\)
\(\,\,\,\,\,1-\frac{4}{x^2}=0\)
\(\,\,\,\,\,1=\frac{4}{x^2}\)
\(\,\,\,\,\,x^2=4\)
\(\,\,\,\,\,x=\pm2\)
\(\,\,\,\,\,f(2)=2+\frac{4}{2}=4\)
\(\,\,\,\,\,f(-2)=-2+\frac{4}{-2}=-4\)
\(\,\,\,\,\,\)The answers are \(y=4\) and \(y=-4\)
The answers are \( y=4 \) and \( y=-4 \)
\(\textbf{14)}\) \( f(x)=\sqrt{x+4} \)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x+4}}\)
\(\,\,\,\,\,\frac{1}{2\sqrt{x+4}}=0\)
\(\,\,\,\,\,\text{A positive fraction cannot equal }0.\)
\(\,\,\,\,\,\text{So there are no horizontal tangent lines.}\)
\(\,\,\,\,\,\)There are no horizontal tangent lines.
There are no horizontal tangent lines.
\(\textbf{15)}\) \( f(x)=\frac{x^2}{x+1} \)
\(\,\,\,\,\,f'(x)=\frac{(x+1)(2x)-x^2(1)}{(x+1)^2}\)
\(\,\,\,\,\,f'(x)=\frac{2x^2+2x-x^2}{(x+1)^2}\)
\(\,\,\,\,\,f'(x)=\frac{x^2+2x}{(x+1)^2}\)
\(\,\,\,\,\,f'(x)=\frac{x(x+2)}{(x+1)^2}\)
\(\,\,\,\,\,x(x+2)=0\)
\(\,\,\,\,\,x=0\text{ or }x=-2\)
\(\,\,\,\,\,f(0)=0\)
\(\,\,\,\,\,f(-2)=\frac{(-2)^2}{-2+1}=-4\)
\(\,\,\,\,\,\)The answers are \(y=0\) and \(y=-4\)
The answers are \( y=0 \) and \( y=-4 \)
\(\textbf{16)}\) If a function has a local minimum, the derivative at that point is positive.
The statement is False.
\(\,\,\,\,\,\text{At a smooth local minimum, the derivative is usually }0.\)
\(\,\,\,\,\,\text{A positive derivative means the function is increasing at that point.}\)
\(\,\,\,\,\,\)The statement is False.
\(\textbf{17)}\) The derivative of a function at a point represents the slope of the tangent line to the graph of the function at that point.
The statement is True.
\(\,\,\,\,\,f'(a)\text{ represents the slope of the tangent line at }x=a.\)
\(\,\,\,\,\,\)The statement is True.
\(\textbf{18)}\) If the derivative of a function is zero at a point, the tangent line to the graph at that point is horizontal.
The statement is True.
\(\,\,\,\,\,f'(a)=0\text{ means the tangent slope is }0.\)
\(\,\,\,\,\,\text{A line with slope }0\text{ is horizontal.}\)
\(\,\,\,\,\,\)The statement is True.
\(\textbf{19)}\) A function with a positive derivative at a point must have a local maximum at that point.
The statement is False.
\(\,\,\,\,\,\text{A positive derivative means the function is increasing at that point.}\)
\(\,\,\,\,\,\text{A local maximum usually happens where the function changes from increasing to decreasing.}\)
\(\,\,\,\,\,\)The statement is False.
\(\textbf{20)}\) The tangent line to a constant function graph is horizontal at every point.
The statement is True.
\(\,\,\,\,\,\text{A constant function has the form }f(x)=c.\)
\(\,\,\,\,\,f'(x)=0\)
\(\,\,\,\,\,\text{Since the slope is always }0,\text{ the tangent line is horizontal everywhere.}\)
\(\,\,\,\,\,\)The statement is True.
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