Question

Find the maxima and minima of function `y=x^3-6x^2+9x-8`?

Answer 1

Maximum at `x=1`.

Minimum at `x=3`.

Explanation:

Given: `y=x^3-6x^2+9x-8`.

We first take the derivative of this function.

`y'=3x^2-12x+9`

Now, we find the zeros of `y'`. Using quadratic formula, we get,

`x=(12+-sqrt(144-4*3*9))/6`

`=(12+-sqrt(36))/6`

`=(12+-6)/6`

`:.x_1=(12+6)/6=3`

`:.x_2=(12-6)/6=1`

Now, we do something called the `ulmathbb("second derivative test")`. It states that

When a function's slope is zero at x, and the second derivative at x is:

• less than 0, it is a local maximum
• greater than 0, it is a local minimum
• equal to 0, then the test fails (there may be other ways of finding out though)

Adapted from: https://www.mathsisfun.com/calculus/maxima-minima.html

So, we now find the second derivative of the function, and get:

`y''=6x-12`

At `x=3`,

`y''=6*3-12`

`=18-12`

`=6`

So, at `x=3`, the function is at its local minimum.

At `x=1`,

`y''=6*1-12`

`=6-12`

`=-6`

So, at `x=1`, the function is at its local maximum.

graph{x^3-6x2-9x-8 [-10, 10, -5, 5]}