Chapter 3.2, Problem 1E

(a)

To determine

The value of $f\text{'}\left(x\right)$

Answer to Problem 1E

  $f\text{'}\left(x\right)=-10x+8$

Explanation of Solution

Given:

The given expression is

  $f\left(x\right)=-5x^2+8x+2;P\left(-1,-11\right)$

Calculation:

The derivative of the function can be calculated as

  $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{-5{\left(x+h\right)}^2+8\left(x+h\right)+2-\left(-5x^2+8x+2\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{-5\left(x^2+h^2+2xh\right)+8x+8h+2+5x^2-8x-2}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{-5x^2-5h^2-10xh+8h+5x^2}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{-5h^2-10xh+8h}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left(-5h-10x+8\right)\\ f\text{'}\left(x\right)=-5\left(0\right)-10x+8\\ f\text{'}\left(x\right)=-10x+8\end{array}$

(b)

To determine

The domain of $f\text{'}\left(x\right)$

Answer to Problem 1E

Set of all the real numbers.

Explanation of Solution

Given:

The given expression is

  $f\left(x\right)=-5x^2+8x+2;P\left(-1,-11\right)$

Calculation:

The domain is nothing but the set of values at which the function is defined. The derivative of the function is

  $f\text{'}\left(x\right)=-10x+8$

The above expression exists for all the real values of x. Therefore, domain of the derivative is the set of all the real numbers.

(c)

To determine

An equation of the tangent line to the graph of ‘f’ at P

Answer to Problem 1E

  $y=18x+7$

Explanation of Solution

Given:

The given expression is

  $f\left(x\right)=-5x^2+8x+2;P\left(-1,-11\right)$

Calculation:

The slope of the function can be obtained by substituting x = -1 in the derivative function,

  $f\text{'}\left(x\right)=-10x+8$

The slope of the function will be

  $\begin{array}{l}f\text{'}\left(-1\right)=-10\left(-1\right)+8\\ f\text{'}\left(-1\right)=10+8\\ f\text{'}\left(-1\right)=18\end{array}$

The equation of the tangent can be calculated as

  $\begin{array}{l}y-\left(-11\right)=18\left(x-\left(-1\right)\right)\\ y+11=18x+18\\ y=18x+18-11\\ y=18x+7\end{array}$

(d)

To determine

The points on the graph at which the tangent line is horizontal

Answer to Problem 1E

  $\left(\frac{4}{5},\frac{26}{5}\right)$

Explanation of Solution

Given:

The given expression is

  $f\left(x\right)=-5x^2+8x+2;P\left(-1,-11\right)$

Calculation:

The tangent line is horizontal if the slope is equal to zero.

Hence, the points at which the tangent line is horizontal will be

  $\begin{array}{l}f\text{'}\left(x\right)=-10x+8=0\\ -10x=-8\\ x=\frac{8}{10}\\ x=\frac{4}{5}\end{array}$

The corresponding y value will be

  $\begin{array}{l}f\left(\frac{4}{5}\right)=-5{\left(\frac{4}{5}\right)}^2+8\left(\frac{4}{5}\right)+2\\ f\left(\frac{4}{5}\right)=\frac{-16}{5}+\frac{32}{5}+2\\ f\left(\frac{4}{5}\right)=\frac{-16+32+10}{5}\\ f\left(\frac{4}{5}\right)=\frac{26}{5}\end{array}$

Hence, the points will be $\left(\frac{4}{5},\frac{26}{5}\right)$