Chapter 2.1, Problem 79E

To determine

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

Answer to Problem 79E

The function $f\left(x\right)=x^2+bx+26$ has minimum value $f_{\min .}\left(x\right)=10$ for $b=\pm 8$ .

Explanation of Solution

Given data:

The function is $f\left(x\right)=x^2+bx+26$ and the given minimum value $f_{\min .}\left(x\right)=10$

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 $f\left(x\right)=x^2+bx+26$ and $f_{\min .}\left(x\right)=10$ .

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

  $\Rightarrow f\left(x\right)=\left(x^2+2\cdot \frac{b}{2}x+\frac{b^2}{4}-\frac{b^2}{4}\right)+26$

  $\Rightarrow f\left(x\right)=\left(x^2+2\cdot \frac{b}{2}x+\frac{b^2}{4}\right)+26-\frac{b^2}{4}$

  $\Rightarrow f\left(x\right)=1\cdot {\left(x+\frac{b}{2}\right)}^2+26-\frac{b^2}{4}$...............(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 $k=26-\frac{b^2}{4}$ .

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

  $\left(h,k\right)=\left(\frac{b}{2},26-\frac{b^2}{4}\right)=\left(h,f_{\min .}\right)$

  $\Rightarrow f_{\min .}=26-\frac{b^2}{4}$ .

But $f_{\min .}=10$ , therefore $26-\frac{b^2}{4}=10$ .

  $\Rightarrow b^2=4\left(26-10\right)=4\times 16=64$

  $\Rightarrow b=\pm \sqrt{64}=\pm 8$

Thus, the function $f\left(x\right)=x^2+bx+26$ has minimum value $f_{\min .}\left(x\right)=10$ for $b=\pm 8$ .

Conclusion:

The function $f\left(x\right)=x^2+bx+26$ has minimum value $f_{\min .}\left(x\right)=10$ for $b=\pm 8$ .