Question

Please, need help! f(x)= (1/24)x^3 -(lnx). need dy/dy and d2y/d2x
to find critical points.

Answer 1

f'(x)=(1/8)x^2-(1/x)
f''(x)=(1/4)x+1/(x^2)

Answer 2

As jahtoday said, now since the derivatives have been computed now set them equal to zero to find critical points. The point of setting them equal to zero is to find where the tangent line is horizontal thus, making them points on the function that "peaks out" at the max and mininum.

Answer 3

set the first derivative equal to zero to find critical points and the second derivative is to find where the points of inflection are. Although this question did not ask that

Answer 4

since ur solving for critical points:
f'(x)=(x^3-8)/(8x)
f''(x)=(x^3+4)/(4x^2)
these form of the equation will make it easier to solve

Answer 5

\[\frac{1}{8}x^2-\frac{1}{x}=\frac{x^3-8}{8x}=0\]

Answer 6

we would say x=0 is a critical number but it doesn't exist in the domain so (0,f(0)) is not a critical point

but (2,f(2)) is a critical point

Answer 7

thanks guys!!! I really appreciate it