Question

Am I doing this right??

Use the second derivative test to locate all relative extrema and inflection points of

f(x)= 2x^4-8x+3

Answer 1

f'(x)= 8x^3-8
f"(x)=24x^2

24x^2=0
sqrt 24 = sqrt x
x= 4.9

Answer 2

yes that's right

Answer 3

@phi what is the next step here to solve the problem?

Answer 4

to find extremes, solve f'(x)=0
8x^3-8 = 0
x^3 = 1
x = 1 (and 2 complex roots)

to find what type of extrema this is, use the 2nd derivative.
f''(x)= 24x^2
at x=1 you get f''(1)= 24. It is positive, which means you found a minimum

Answer 5

ok got it . thanks!

Answer 6

@phi I'm sorry, one more question? how do i get the inflection points again?

Answer 7

inflection points, max pts and min pts all have 0 slope i.e. 1st derivative is zero at those types of points.

here there is only 1 point where f'(x) =0. It occurs at x=1.
According to the 2nd derivative test, it is a minimum

Answer 8

So The only inflection point is x=1

Answer 9

Thank you!

Answer 10

x=1 is not an inflection point. x=1 is a minimum

see http://mathworld.wolfram.com/InflectionPoint.html