Question

f(x)=x^5-5x^4+60 Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither.

Answer 1

I think the f'(x)=5x^4-20x^3 and f"(x)=20x^3-60x^2 but i dont know where to go from there

Answer 2

Critical points are where f' = 0. Set it equal to 0, solve for x.

Inflection points are where f''' = 0. Do as above^^

Answer 3

so I got critical points as 0 and 4?

Answer 4

\[\large 0=5x^3(x-4)\]yep

Answer 5

f"(x)=20x^3-60x^2
\[\Large 0 = 20x^2(x-3)\]

Answer 6

that one would be for the inflection right?

Answer 7

so the f"(0) =0 and itd be neither and f"(4)=320 and that'd be min right?

Answer 8

You got both inflection points? x = 0 and x=...?

Answer 9

Actually x=0 is a local max, https://www.google.com/search?q=x%5E5-5x%5E4%2B60&oq=x%5E5-5x%5E4%2B60&aqs=chrome..69i57&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8

Answer 10

the other inflection point is 3 right? but why wouldn't you use the 0 as one? (my online homework only allows one answer to be put in which im assuming is 3).

Answer 11

Yes, since x=0 isn't an inflection point, it's a turning point. Concavity doesn't change at a turning point.

Answer 12

okay and then one more question, why would the critical point of 0 be a max point and not either?

Answer 13

You can show it's a maximum with the first derivative test - to the left of x=0, slope is positive. To the right, slope is negative... thus it's a maximum.

Answer 14

The 2nd derivative being 0 doesn't guarantee an inflection point.

Answer 15

At an inflection point, the slope doesn't change sign, if it's a positive slope before it, it's still positive after it.