Question

find all relative extrema by using the Second Derivative test for f(x)=x^(4)-4x^(3)+2

i have found the what the first derivative test which is the square root of _+ 2 and 0 and now i am figuring for the second derivative and trying to prove if it works on the second derivative please help

Answer 1

f(x) = x^4 - 4x^3 + 2
Find all critical points:
f'(x) = 4x^3 - 12x^2 = 0
4x^2(x - 3) = 0
x = 0, 3 are critical points.
Find the second derivative and test the critical points:
f''(x) = 12x^2 - 24x = 12x(x-2)
Find f''(0) and f''(3)

Answer 2

f''(0) = 0. Therefore, x = 0 is a point of inflection.
f''(3) = 12*3*1 which is positive and therefore x = 3 is a minima.
The minimum value is: f(3) = 3^4 - 4(3)^3 + 2 = 81 - 108 + 2 = -25.

Answer 3

ok thank you so much

Answer 4

You are welcome.

Answer 5

x = 0 is a point of inflection and therefore it is not considered to be an extrema (neither minima nor maxima). There is only one extrema at x = 3 and it is a minima.

Answer 6

so there for the first derivative test fails and you have to do the second derivative test right?

Answer 7

First derivative when equated to zero will identify all critical points.
Then we have to use the second derivative test to test each critical point.
If the second derivative is positive it is a minima.
If the second derivative is negative it is a maxima.
If the second derivative is zero it is a point of inflection.

Answer 8

So it is not a question of the first derivative test failing.
First derivative identifies critical points.
Second derivative helps classify the critical points as: minima, maxima or point of inflection.

Answer 9

ok thank you

Answer 10

You are welcome.