Given f(x) = 3x^4 - 2x^3 + 4, find f', intervals of incr/dec

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Jade

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Sep 16, 2006
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Part A) I need to find f'(x) which I have 12x^3-6x^3

Then find the critical numbers which I came up with x=0,1/2

Part B) then I need to find the interval(s) where f(x) is increasing - I came up with that it is increasing throughout. I plugged in test numbers before 0, in between 0 and 1/2 and after 1/2 it is still increasing. What would the intervals be?

Part C) Then I am asked the intervals that it is decreasing which I believe is none?

Part D) I am asked to find all relative minimum and relative maximum if any. Give both x and y values. I believe there is not one because it keeps increasing.

Part E) Find all points of inflection. I know you need to find f''(x) which is 36-12x then set it to zero. Which in this case x=3 (I would say POI is (3, 12)), but what does that mean because I thought there isn't a POI if it keeps increasing.

Part F) Where is f(x) concave up. I understand that concave up is when it looks like a cup. Do I need to test more points to help me with this???

Finally Part G) Draw a short arc around the point (-1,9). I feel that I have made a mistake somewhere in the beginning problems to get this far. Am I on the right track with this question????
 
Re: Given f(x)=3x^4-2x^3+4

Hello, Jade!

Are you sure that the function is: \(\displaystyle \,f(x)\:=\:12x^3\,-\,6x^3\) ??

 
The function

The function is
f(x)=3x^4-2x^3+4, but then I am suppose to take the derivative of that function which is f'(x)=12x^3-6x^2 - oops I think I wrote it wrong the first time.
 
f(x) = 3x^4 - 2x^3 + 4

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

12x^3 - 6x^2 = 0

6x^2(2x - 1) = 0

x = 0, x = 1/2

for x < 0, f'(x) < 0 ... f(x) is decreasing

for 0 < x < 1/2, f'(x) < 0 ... f(x) is decreasing

for x > 1/2, f'(x) > 0 ... f(x) is increasing

absolute min at x = 1/2

f"(x) = 36x^2 - 12x

36x^2 - 12x = 0

12x(3x - 1) = 0

x = 0 or x = 1/3

for x < 0, f"(x) > 0 ... f(x) is concave up

for 0 < x < 1/3, f"(x) < 0 ... f(x) is concave down

for x > 1/3, f"(x) > 0 ... f(x) is concave up

you now know where inflection points are located, correct?
 
Question

Before I answer that I have a question. When you find your critical numbers 0 and 1/2 do you input them into the original function or the derivative? I am thinking the original function. Is this right? Also when you find the second derivative and get the critical numbers of 0 and 1/3 do you insert them into the original function as well? I want to be clear before I go any further. I am really excited that we both got the same answer for the critical numbers :D
 
If I plug in the x then I should get y in the original f(x)

critical numbers are x=0,1/3

If I plug in x=0 then I get y=4, one point of inflection would be (0,4)

If I plug in x=1/3 then I get y=3.96, the other POI would be (1/3, 3.96)
 
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