Chapter 2.4, Problem 34E

To determine

To find: The point at which the tangent to the curve $f\left(x\right)=3-4x-x^2$ is horizontal.

Answer to Problem 34E

The point at which the tangent to the curve $f\left(x\right)=3-4x-x^2$ is horizontal is $\left(-2,7\right)$.

Explanation of Solution

Given information:

The curve is $f\left(x\right)=3-4x-x^2$.

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.

  $\begin{array}{l}f\left(x\right)=3-4x-x^2\\ f\left(a+h\right)=3-4\left(a+h\right)-{\left(a+h\right)}^2\\ f\left(a+h\right)=3-4a-4h-\left(a^2+h^2+2ah\right)\\ f\left(a+h\right)=3-4a-a^2-h\left(4+2a\right)-h^2\end{array}$

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

  $\begin{array}{l}f\left(x\right)=3-4x-x^2\\ f\left(a\right)=3-4\left(a\right)-{\left(a\right)}^2\\ f\left(a\right)=3-4a-a^2\end{array}$

Substitute $3-4a-a^2-h\left(4+2a\right)-h^2$ for $f\left(a+h\right)$ and $3-4a-a^2$ for $f\left(a\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{3-4a-a^2-h\left(4+2a\right)-h^2-\left(3-4a-a^2\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{-h\left(4+2a\right)-h^2}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(-4-2a-h\right)}{h}\\ =\underset{h\to 0}{\lim }\left(-4-2a-h\right)\end{array}$

Further simplify.

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

Now, equate the slope to zero.

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

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

  $\begin{array}{l}f\left(x\right)=3-4x-x^2\\ f\left(-2\right)=3-4\left(-2\right)-{\left(-2\right)}^2\\ f\left(-2\right)=3+8-4\\ f\left(-2\right)=7\end{array}$

Therefore, the point at which the tangent to the curve $f\left(x\right)=3-4x-x^2$ is horizontal is $\left(-2,7\right)$.