Chapter SH, Problem 9.1.17EP

(a)

To determine

Whether the function has maximum or minimum value.

Answer to Problem 9.1.17EP

The function has a maximum value.

Explanation of Solution

Given:

The given function is:

  $y=-x^2+x+1$

Calculation:

Finding the first and second derivatives of the given function:

  $\begin{array}{l}y=-x^2+x+1\\ \frac{dy}{dx}=-2x+1\\ \frac{d^{2}y}{d^{2}x}=-2\\ Also,\\ -2x+1=0\\ 2x=0\\ x=\frac{1}{2}\end{array}$

Calculating the value of $y=-x^2+x+1$ .

  $\begin{array}{l}y=-{\left(\frac{1}{2}\right)}^2+\left(\frac{1}{2}\right)+1\\ y=-\frac{1}{4}+\frac{3}{2}\\ y=\frac{-1+6}{4}\\ y=\frac{5}{4}\end{array}$

The second derivative of the given function is negative. Hence, the function has a maximum value.

(b)

To determine

The maximum or minimum value.

Answer to Problem 9.1.17EP

  $\left(\frac{1}{2},\frac{5}{4}\right)$

Explanation of Solution

Given:

The given function is:

  $y=-x^2+x+1$

Calculation:

The maximum value is:

  $\left(\frac{1}{2},\frac{5}{4}\right)$

(c)

To determine

The domain and range of the given function.

Answer to Problem 9.1.17EP

The domain will be the set of all real numbers and range will be $\{y:y\le \frac{5}{4}\}.$

Explanation of Solution

Given:

The given function is:

  $y=-x^2+x+1$

Calculation:

The graph of the given function will be:

  

The domain will be the set of all real numbers and range will be $\{y:y\le \frac{5}{4}\}.$