Chapter 4.7, Problem 64SR

To determine

  $f\left(x\right)=-4x^2+x-18$

Whether the function has maximum or minimum value.

Answer to Problem 64SR

The function has a maximum value and the value is $-\frac{287}{16}$ .

Explanation of Solution

Given information:

The function mentioned in the problem is $f\left(x\right)=-4x^2+x-18$ .

Formula used:

For the maximum value of the function $f\left(x\right)$ the second derivative of the function $f\left(x\right)$ should be negative value $i.e,\left(<0\right)$ .

For the minimum value of the function $f\left(x\right)$ the second derivative of the function $f\left(x\right)$ should be positive value $i.e,\left(>0\right)$ .

For the critical point the first derivative of the function $f\left(x\right)$ should be equal to zero and at this point we will get the maximum or minimum value of the function $f\left(x\right)$ .

Calculation:

Consider that the function $f\left(x\right)=-4x^2+x-18$ .

Recall that the for the maximum value of the function $f\left(x\right)$ the second derivative of the function $f\left(x\right)$ should be negative value $i.e,\left(<0\right)$ and for the minimum value of the function $f\left(x\right)$ the second derivative of the function $f\left(x\right)$ should be positive value $i.e,\left(>0\right)$ .

Therefore the first derivative of the function $f\left(x\right)=-4x^2+x-18$ is $f\text{'}\left(x\right)=-8x+1$ and the second derivative of function $f\left(x\right)=-4x^2+x-18$ is $f\text{'}\text{'}\left(x\right)=-8$ which is less than zero.

Therefore the function has maximum value.

Recall that for the critical point the first derivative of the function $f\left(x\right)$ should be equal to zero and at this point we will get the maximum or minimum value of the function $f\left(x\right)$ .

Therefore the first derivative is equal to zero than we have $f\text{'}\left(x\right)=0\Rightarrow -8x+1=0\Rightarrow x=\frac{1}{8}$ is the critical point.

Therefore at $x=\frac{1}{8}$ the function $f\left(x\right)=-4x^2+x-18$ has the value $f\left(x\right)=f\left(\frac{1}{8}\right)=-4*{\left(\frac{1}{8}\right)}^2+\left(\frac{1}{8}\right)-18=-4*\frac{1}{64}+\frac{1}{8}-18=-\frac{1}{16}+\frac{1}{8}-18=-\frac{287}{16}$ .

Therefore the function $f\left(x\right)=-4x^2+x-18$ has minimum value and the value is $-\frac{287}{16}$ .