Question

What are the local extrema of `f(x)= -x^3 + 3x^2 + 10x + 13`?

Answer 1

Local maximum is `25 + (26sqrt(13/3))/3`
Local minimum is `25 - (26sqrt(13/3))/3`

Explanation:

To find local extrema, we can use the first derivative test. We know that at a local extrema, at the very least the function's first derivative will equal zero. So, let's take the first derivative and set it equal to 0 and solve for x.

`f(x) = -x^3 + 3x^2 + 10x +13`

`f'(x) = -3x^2 + 6x + 10`

`0 = -3x^2 + 6x + 10`

This equality can be solved easily with the quadratic formula. In our case, `a = -3`, `b = 6` and `c=10`

Quadratic formula states:

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

If we plug back our values into the quadratic formula, we get
`x = (-6 +- sqrt(156))/-6 = 1 +- sqrt(156)/6 = 1 +- sqrt(13/3)`

Now that we have the x values of where the local extrema are, let's plug them back into our original equation to get:

`f(1+sqrt(13/3)) = 25 + (26sqrt(13/3))/3` and

`f(1 - sqrt(13/3)) = 25 - (26sqrt(13/3))/3`