On what interval is h(x) concave down?

Question

The derivative of function h is given by h'(x) = x(x-2)^3
On what interval is h(x) concave down?
A) x<.5
B) x>.5
C) 0<x<2
D)1<x<2
E) x>2

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Answer 1 · 2018-02-03T08:27:42

#A) \ x<0.5# should be the correct answer.

We have #h'(x)=x(x-2)^3#

Let's expand the function first.

#h'(x)=x(x-2)^3#

#h'(x)=x(x^3-6x^2+12x-8)#

#h'(x)=x^4-6x^3+12x^2-8x#

Now, let's find the derivative of this function to get the second derivative of the original function.

#:. h''(x)=4x^3-18x^2+24x-8#

If you want #h(x)# to be concave down, then the conditions are: #h''(x)<0#.

Now, we have

#4x^3-18x^2+24x-8<0#

We can factor the #LHS# to get

#2(2x-1)(x-2)^2<0#

We have either #(2x-1)<0# or #(x-2)^2<0#

So, either

#x<1/2# or #x<2#

Combining, we get

#x<1/2<2#

#:.x<1/2#

From here, we can clearly see that option #A) \ x<0.5# is correct.