Consider the function 𝑓(𝑥) = 12𝑥^5 + 30𝑥^4 − 160𝑥^3 + 4

Question

a. Find the inflection points.

b. Find the intervals where the graph of the function is concave up.

c. Find the intervals where the graph of the function is concave down.

When I first tried this problem, I got very complicated answers that did not seem correct.

Answer 1

Concavity is determined by the 2nd derivative.

So first, you need to take the derivative once and then again to get the second derivative.

Then the 2nd derivative, set it equal to 0. This will give you the boundary values or critical values.

Plug in any number from each interval or number between two critical values.

When you plug a number for x into the 2nd derivative; if the result is positive then it is concave up, if the result is negative then it is concave down.

When a value of x plugged into the 2nd derivative is equal to 0, that is an inflection point. An inflection point is the transition from concave up to concave down or a transition from concave down to concave up.

So your original function is 12x⁵+30x⁴-160x³+4

The 1st derivative is 60x⁴+120x³-480x²

And the 2nd derivative is 240x³+360x²-960x

Now set the 2nd derivative equal to 0

240x³+360x²-960x = 0 and solve for all values of x.

so factor out an x, since each term has at least one x.

In that case x (240x²+360x-960) = 0

Well x = 0 is one answer. Now use quadratic formula to get the other 2 values for x.

I used a calculator, forgive me, but I get x = -2.886 and x = 1.386

So the 3 roots are -2.886, 0, and 1.386

Now pick a value to test each interval

For example, plug in x = -10 for the 2nd derivative, you get a negative value, this interval from negative infinity to -2.886 is concave down.

Now plug in x = -1 for the 2nd derivative, you get a positive value, this interval from -2.886 to 0 is concave up.

Now plug in x = 1 for the 2nd derivative, you get a negative value, this interval from 0 to 1.386 is concave down.

Finally plug in x = 2 for the 2nd derivative, you get a positive value, this interval from 1.386 to positive infinity is concave up.

x = -2.886, x = 0, and x = 1.386 are inflection points, but you should plug those into the original equation to verify that the value of the original function is not undefined. In this case, since there is no denominator the original function is not a concern and all 3 values of x are inflection points.

Hope this helps.