Chapter 2.4, Problem 3QQ

To determine

Tofind:The correct choice for an equation of the tangent line to $f\left(x\right)=9-x^2$ at $x=2$ .

Answer to Problem 3QQ

The correct choice is (B) such that the equation of the tangent line to $f\left(x\right)=9-x^2$ at $x=2$ is $y=-4x+13$ .

Explanation of Solution

Given information:The curve is $f\left(x\right)=9-x^2$ and the point of tangency is $x=2$ .

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

Substitute $2+h$ for $x$ in the function $f\left(x\right)=9-x^2$ .

  $\begin{array}{c}f\left(2+h\right)=9-{\left(2+h\right)}^2\\ =9-\left(4+h^2+4h\right)\\ =5-h^2-4h\end{array}$

Substitute $2$ for $x$ in the function $f\left(x\right)=9-x^2$ .

  $\begin{array}{c}f\left(2\right)=9-{\left(2\right)}^2\\ =9-4\\ =5\end{array}$

Substitute the values of $f\left(2+h\right)$ and $f\left(2\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{5-h^2-4h-5}{h}\\ =\underset{h\to 0}{\lim }\frac{-h\left(h+4\right)}{h}\\ =\underset{h\to 0}{\lim }\left(-\left(h+4\right)\right)\end{array}$

Further simplify.

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

So, the slope of the tangent line is $-4$ .

The point of tangency to the curve $f\left(x\right)$ is $\left(a,f\left(a\right)\right)$ that is $\left(2,f\left(2\right)\right)=\left(2,5\right)$ .

The equation of the line passing through $\left(x_1,y_1\right)$ and slope $m$ is $y-y_1=m\left(x-x_1\right)$ .

Substitute 2 for $x_1$ , $5$ for $y_1$ and $-4$ for $m$ in the equation of line.

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-5=\left(-4\right)\left(x-2\right)\\ y-5=-4x+8\\ y=-4x+13\end{array}$

Therefore, the correct choice is (B) such that the equation of the tangent line to $f\left(x\right)=9-x^2$ at $x=2$ is $y=-4x+13$ .