Chapter 4.7, Problem 65SR

To determine

Whether the function has maximum or minimum value.

Answer to Problem 65SR

The function $f\left(x\right)=3x^2-9+6x$ has a minimum value and the value is $-12$ .

Explanation of Solution

Given information:

The function mentioned in the problem is $f\left(x\right)=3x^2-9+6x$ .

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 $f\left(x\right)=3x^2-9+6x$ .

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 $f\left(x\right)=3x^2-9+6x$ is $f\text{'}\left(x\right)=6x-0+6=6x+6$ and the second derivative of function $f\left(x\right)=3x^2-9+6x$ is $f\text{'}\text{'}\left(x\right)=6+0=6$ which is greater than zero.

Therefore the function has minimum 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 $f\text{'}\left(x\right)=0\Rightarrow 6x+6=0\Rightarrow x=-1$ is the critical point.

Therefore at $x=-1$ the function $f\left(x\right)=3x^2-9+6x$ has the value $f\left(x\right)=f\left(-1\right)=3*{\left(-1\right)}^2-9+6\left(-1\right)=3-9-6=-12$ .

Therefore the function $f\left(x\right)=3x^2-9+6x$ has minimum value and the value is $-12$ .