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

1 Answer
Nov 4, 2015

Answer:

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#