Chapter 4.7, Problem 63SR

To determine

Whether the function has maximum or minimum value.

Answer to Problem 63SR

The function $f\left(x\right)=6x^2-8x+12$ has a minimum value and the value is $\frac{28}{3}$ .

Explanation of Solution

Given information:

The function mentioned in the problem is.

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 $f\left(x\right)=6x^2-8x+12$ .

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 $f\left(x\right)=6x^2-8x+12$ is $f\text{'}\left(x\right)=12x-8$ and the second derivative of function $f\left(x\right)=6x^2-8x+12$ is $f\text{'}\text{'}\left(x\right)=12-0=12$ 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 $f\text{'}\left(x\right)=0\Rightarrow 12x-8=0\Rightarrow x=\frac{8}{12}$ is the critical point.

Therefore at $x=\frac{8}{12}$ the function $f\left(x\right)=6x^2-8x+12$ has the value $f\left(x\right)=f\left(\frac{8}{12}\right)=6*{\left(\frac{8}{12}\right)}^2-8\left(\frac{8}{12}\right)+12=6*\frac{64}{144}-8*\frac{8}{12}+12=\frac{8}{3}-\frac{16}{3}+12=\frac{28}{3}$ .

Therefore the function $f\left(x\right)=6x^2-8x+12$ has minimum value and the value is $\frac{28}{3}$ .