Chapter 2.4, Problem 33E

To determine

To find: The point at which the tangent to the curve $f\left(x\right)=x^2+4x-1$ is horizontal.

Answer to Problem 33E

The point at which the tangent to the curve $f\left(x\right)=x^2+4x-1$ is horizontal is $\left(-2,-5\right)$.

Explanation of Solution

Given information:

The curve is $f\left(x\right)=x^2+4x-1$.

Calculation:

The tangent to any point is horizontal if the slope of the curve at that point is equal to $0$.

The formula for the slope of curve at $x=a$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

Substitute $a+h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2+4x-1\\ f\left(a+h\right)={\left(a+h\right)}^2+4\left(a+h\right)-1\\ f\left(a+h\right)=\left(a^2+h^2+2ah\right)+4a+4h-1\\ f\left(a+h\right)=h^2+\left(2a+4\right)h+a^2+4a-1\end{array}$

Substitute $a$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2+4x-1\\ f\left(a\right)={\left(a\right)}^2+4\left(a\right)-1\\ f\left(a\right)=a^2+4a-1\end{array}$

Substitute $h^2+\left(2a+4\right)h+a^2+4a-1$ for $f\left(a+h\right)$ and $a^2+4a-1$ for $f\left(a\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{h^2+\left(2a+4\right)h+a^2+4a-1-\left(a^2+4a-1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{h^2+2ah+4h}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(h+2a+4\right)}{h}\\ =\underset{h\to 0}{\lim }\left(h+2a+4\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=0+2a+4\\ =2a+4\end{array}$

Now, equate the slope to zero.

  $\begin{array}{c}2a+4=0\\ 2a=-4\\ a=\frac{-4}{2}\\ a=-2\end{array}$

Substitute $-2$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2+4x-1\\ f\left(-2\right)={\left(-2\right)}^2+4\left(-2\right)-1\\ f\left(-2\right)=4-8-1\\ f\left(-2\right)=-5\end{array}$

Therefore, the point at which the tangent to the curve $f\left(x\right)=x^2+4x-1$ is horizontal is $\left(-2,-5\right)$.