Chapter 12.3, Problem 10E

To determine

To find: The slope of the graph of the function at the specified point.

Answer to Problem 10E

The slope of the graph of the function at the specified point is $-2$ .

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=10x-2x^2$

The given point is $\left(3,12\right)$ .

Concept used:

The slope of the tangent line to the graph at $\left(x,f\left(x\right)\right)$ is equal to the slope of its tangent line at $\left(x,f\left(x\right)\right)$ and is given by,

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

The given function is,

  $f\left(x\right)=10x-2x^2$

The slope of the graph of the function at the specified point is calculated as,

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[10\left(x+h\right)-2{\left(x+h\right)}^2\right]-\left[10x-2x^2\right]}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[\left(10x+10h\right)-2\left(x^2+h^2+2xh\right)\right]-\left[10x-2x^2\right]}{h}\\ =\underset{h\to 0}{\lim }\frac{10h-2h^2-4xh}{h}\end{array}$

Simplify further,

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{h\left(10-2h-4x\right)}{h}\\ =-4x+10\\ =-4\left(3\right)+10\\ =-2\end{array}$

Consider the graph of the given function as,

  

Therefore, the slope of the graph of the function at the specified point is $-2$ .