Chapter 2.1, Problem 80E

To determine

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

Answer to Problem 80E

The function $f\left(x\right)=x^2+bx-25$ has minimum value $f_{\min .}\left(x\right)=-50$ for $b=\pm 10$ .

Explanation of Solution

Given data:

The function is $f\left(x\right)=x^2+bx-25$ and the given minimum value $f_{\min .}\left(x\right)=-50$

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 upwards, and minimum value of function is at the vertex, i.e. vertex is $\left(h,k\right)=\left(h,f_{\min .}\right)$

The givenfunction is $f\left(x\right)=x^2+bx-25$ and $f_{\min .}\left(x\right)=-50$ .

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

  $\Rightarrow f\left(x\right)=\left(x^2+2\cdot \frac{b}{2}x+\frac{b^2}{4}-\frac{b^2}{4}\right)-25$

  $\Rightarrow f\left(x\right)=\left(x^2+2\cdot \frac{b}{2}x+\frac{b^2}{4}\right)-25-\frac{b^2}{4}$

  $\Rightarrow f\left(x\right)=1\cdot {\left(x+\frac{b}{2}\right)}^2-\left(25+\frac{b^2}{4}\right)$...............(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 $k=-\left(25+\frac{b^2}{4}\right)$ .

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

  $\left(h,k\right)=\left(\frac{b}{2},-\left(25+\frac{b^2}{4}\right)\right)=\left(h,f_{\min .}\right)$

  $\Rightarrow f_{\min .}=-\left(25+\frac{b^2}{4}\right)$ .

But $f_{\min .}=10$ , therefore $-\left(25+\frac{b^2}{4}\right)=-50$ .

  $\Rightarrow 25+\frac{b^2}{4}=50$

  $\Rightarrow b^2=4\left(50-25\right)=100$

  $\Rightarrow b=\pm \sqrt{100}=\pm 10$

Thus, the function $f\left(x\right)=x^2+bx-25$ has minimum value $f_{\min .}\left(x\right)=-50$ for $b=\pm 10$ .

Conclusion:

The function $f\left(x\right)=x^2+bx-25$ has minimum value $f_{\min .}\left(x\right)=-50$ for $b=\pm 10$ .