Question

Consider the function f(x)= 3/4x^4-x^3-3x^2+6x

Answer 1

1. Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values.

2. Determine the interval(s) where f(x) is increasing (if any) and the interval(s) where f(x) is decreasing (if any).

3. Show why there are exactly two inflection points for this function. Note: You do NOT need to find the inflection points.

4. D. Show f(x) is concave up at x = −2, x = −1, and x = 2. And show f(x) is concave down at x = 0. Use this information and the information in parts A–C to sketch this function.

Answer 2

Is that whole portion in the denominator of the one fraction?

Answer 3

\[f(x) = (\frac{ 3 }{ 4 })^{4}-x^{3}-3x ^{2}+6x \]

Answer 4

With a missing x in there Im assuming. Alrighty. So we want extrema, which means finding thefirst derivative and setting it equal to 0 first. Can you get the first derivative okay?

Answer 5

sorry, it should be (3/4)x^4

Answer 6

I know, no worries. Just need thefirst derivative for now :3

Answer 7

f'(x) = 81x^3/64 - 3x^2 -6x +6
Right?

Answer 8

You didnt need to raise the 3 and the 4 to a power, the way its written only x should be raised to the 4th power.

Answer 9

Sorry, I'm not quite understanding what you mean.

Answer 10

(3/4)x^4 just becomes 3x^3, but you took the 3/4 fraction to the 4th power as well when you didnt need to.

Answer 11

f'(x) =-3x^2+6x +6

Answer 12

And then we find the second derivative, right?

Answer 13

Yeah, we have 3x^3 - 3x^2 + 6x + 6. But before we jump to the second derivative, we need the extrema, which means we set this derivative equal to 0 and solve for x.

Answer 14

Ahh, this is kind of complicated. I got three answers?

Answer 15

Well, you can solve for x in f'(x) by factoring by grouping.
(3x^3-3x^2) + (-6x+6)
3x^2(x - 1) -6(x-1)
(3x^2-6)(x-1)
x = 1
x = +/- sqrt(2)

Answer 16

When I factored, I had gotten 3(x^3-x^2+2x+2). Sorry, math is not my forte.

Answer 17

No worries. Either way, it turns out the same. I just noticed it as a grouping factoring.

Answer 18

Okay, so now do we plug in +/- sqrt(2) into the second derivative?

Answer 19

Nope, no need. So we have these as extrema. Now we need increasing and decreasingintervals, though. This means weneed to pick numbers on each side of those 3 answers. When we choose a number in each side, we plug it into the first derivative and check the sign of the answer. if the sign is negative, we are decreasing on that side of our extrema. If we get a positive answer then we areincreasing on that side of the extrema.