How do you find the exact relative maximum and minimum of the polynomial function of #f(x) =2x^3-3x^2-12x#?

1 Answer
May 2, 2018

Maximum #=(-1,7)#
Minimum #=(2,-20)#

Explanation:

Since the graph is not limited by a definition, we can just look for the extremal points through the dervivate of the function.

#2x^3-3x^2-12x#

Dervivating the function with rule #(x^n)'=nx^(n-1)#, will give.

#(2x^3-3x^2-12x)'=6x^2-6x-12#

After finding the #f'(x)# we can look for zero points of that graph. Because this will be equal to the extremal points.

#6x^2-6x-12=0#

From here we can take several paths. Such as either use the quadratic equation or factorize to find the zero points. This will give us.

#x=2# and #x=-1# which can be seen as #(x-2)(x+1)#.

We can now put these numbers into the original function #f(x)# and find the #y-#values of the extremal points.

#f(2)=2*2^3-3*2^2-12*2=-20#
#f(-1)=2*(-1)^3-3*(-1)^2-12*(-1)=7#

These will be equal to coordinates, and we can with logic say which is which by this calculation. Since #f(2)=-20# we know that this has to be equal to the minimum point. And that #f(-1)=7# has to be equal to the maximum point.

So by that, we can say that.

Maximum #=(-1,7)#
Minimum #=(2,-20)#