Chapter 2.1, Problem 82E

To determine

To find:The value of b such that the function has the given maximum value.

Answer to Problem 82E

The required value of b is $b=\pm 16$

Explanation of Solution

Given data:

The function is $f\left(x\right)=-x^2+bx-16$ and the given maximum value $f_{\max .}\left(x\right)=48$

Concept used:

The standard form of quadratic function $f\left(x\right)=a{\left(x-h\right)}^2+k$ represents a parabola with vertex $\left(h,k\right)$ . If $a<0$ then the graph of quadratic function opens downwards, and maximum value of function is at the vertex, i.e. vertex is $\left(h,k\right)=\left(h,f_{\max .}\right)$

The givenfunction is $f\left(x\right)=-x^2+bx-16$ and $f_{\max .}\left(x\right)=48$ .

The function $f\left(x\right)$ may be written as

  $\Rightarrow f\left(x\right)=-\left(x^2-2\cdot \frac{b}{2}x+\frac{b^2}{4}-\frac{b^2}{4}\right)-16$

  $\Rightarrow f\left(x\right)=-\left(x^2-2\cdot \frac{b}{2}x+\frac{b^2}{4}\right)+\frac{b^2}{4}-16$

  $\Rightarrow f\left(x\right)=-1\cdot {\left(x-\frac{b}{2}\right)}^2+\frac{b^2}{4}-16$...............(1)

This is standard form of the quadratic function $f\left(x\right)$ . Now, comparing result (1) with $f\left(x\right)=a{\left(x-h\right)}^2+k$ . We have $a=-1$ , $h=\frac{b}{2}$ , and $k=\frac{b^2}{4}-16$ .

Since, $a=-1<0$ , so the function $f\left(x\right)$ concave downward and so its vertex is

  $\left(h,k\right)=\left(\frac{b}{2},\frac{b^2}{4}-16\right)=\left(h,f_{\max .}\right)$

  $\Rightarrow f_{\max .}=\frac{b^2}{4}-16$ .

But $f_{\max .}\left(x\right)=48$ , therefore $\frac{b^2}{4}-16=48$ .

  $\Rightarrow b^2=4\left(48+16\right)=4\times 64$

  $\Rightarrow b=\pm \sqrt{4\times 64}=\pm 16$

Thus, the function $f\left(x\right)=-x^2+bx-16$ has maximum value $f_{\max .}\left(x\right)=48$ for $b=\pm 16$ .

The answer can be verified from figure given below.

  

Thus, for value $b=\pm 16$ the function $f\left(x\right)=-x^2+bx-16$ has maximum value 48.

Conclusion:

For $b=\pm 16$ the function $f\left(x\right)=-x^2+bx-16$ has maximum value 48.