What are the values of a and b if f(x)=ax^2+b/x^3 has extreme value at (3,2)?

Question

1715 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-10T17:13:03

#a = 2/15#
#b =108/5#

We find the derivative:

#f'(x) = 2ax - 3b/x^4#

If there is an extreme value, then the derivative equals #0# at that point.

#0 = 2a(3) - (3b)/(3^4)#

#b/27 = 6a#

#b = 162a#

Now we also know that the function #f(x)# must pass through #(3, 2)#.

#2 = a(3)^2 + b/3^3#

#2 = 9a + b/27#

#2 = 9a + (162a)/27#

#2 = 9a + 6a#

#2 = 15a#

#a = 2/15#

Therefore:

#b = 162(2/15) = 21.6#

Let's try to graph this and see if it makes sense.

enter image source here

As you can see, there is indeed an extreme value at #(3, 2)#.

Hopefully this helps!