Local Min and Max values?

Question

Let f(x)=x^6+16x^3. Find the local minimum and maximum points and values of the function f(x).

This problem was on my third calculus test and I got half of it correct, but I believe in actuality that I got the entire problem right.

I came up with two critical values of (x=0, x=-2)

Local Minimum at x=-2, with a value of -64
Local Maximum at x=0, with a value of 0

Could someone please double check this problem for me to make sure that it is in fact incorrect.

Thank You

Answer 1

f' = 6x^5 + 16*3x^2 = 0 = 6x^2(x^3 + 8)

crit val... 0 & -2 ... correct...

note that 6x^2 (x+2)(x^2 - 2x + 4)
only x+2 will change sign...
the others are always nonnegative...

thus f' < 0 if x < -2
f' > 0 if x > -2...

thus minimum at -2...
there is no local maximum (it turns out that 0 is an inflection...)


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Answer 2

Hello

This function approaches infinity as x decreases without bound and approaches infinity as x increases without bound. Hence, it can not have a local min at x=-2 and a max at x = 0 without having another min somewhere.

Look at the second derivative to assertain which ones are mins and maxes and where the points of inflection are.

Hope This Helps!!

Answer 3

Your local min is correct, but I believe that x=0 is simply an inflection point, not a local max. Sorry.

Answer 4

Well, from what you've given above it appears to be ok, but without knowing exactly how the problem was stated on the exam I can't confirm or deny that you should get more points. You should really take this up with whoever graded your exam.

Answer 5

f(x)=x^6+16x^3
f'(x)=6x^5+48x^2

6x^5+48x^2 = 0
6x^3+48 = 0
x^3+8= 0
x^3 = -8

Looks like only a min/max at -2

Answer 6

Did you have to find the points on which the function was increasing or decreasing?

Answer 7

Your critical numbers are right; if you got the Q wrong, it was b/c of substitution, which I don't really want to do at this time.