Question

find the extreme values of the following y = X3 - 6X2 + 9X - 8

Answer 1

Let's first find the derivative of the function:

$$\dfrac{dy}{dx}=3x^2-12x+9.$$

Function have the extreme values when $\dfrac{dy}{dx}=0$:

$$3x^2-12x+9=0.$$

This quadratic equation has two roots: $x_1=3,$ $x_2=1$. Therefore, the extreme values of the function are: $x=1$ and $x=3$. Let's also find the nature of the extreme values. Let’s divide the function into the following intervals: $x<1$, $1<x<3,$ $x>3$. Then, we can choose the test points in these intervals: $x=0$, $x=2$ and $x=4.$ Finally, we can determine the sign of the derivative in these intervals:

$$f'(0)=3\cdot(0)^2-12\cdot0+9=9>0,$$ $$f'(2)=3\cdot(2)^2-12\cdot2+9=-3<0,$$ $$f'(4)=3\cdot(4)^2-12\cdot4+9=9>0.$$

As we can see from calculations, derivative has sign plus on interval $x<1$ and sign minus on interval $1<x<3$. Since derivative change its sign from positive to negative, the extreme value at point $x=1$ is the maximum. Also, we can see that derivative has sign plus on interval $x>3$. Since derivative change its sign from negative to positive, the extreme value at point $x=3$ is the minimum.