Question

Help please!

Find the critical numbers of the function f(t)=6t^(2/3)+t^(5/3)

there are two critical numbers.

Answer 1

ok... so whats the 1st derivative..?

and you should also find the 2nd derivative as well

Answer 2

The derivative of the function would be \[((5x^(2/3)/(3))+((4)/(x^(1/3))\] how do I find the second derivative? @campbell_st

Answer 3

wow... I thought

\[f'(t) = 6 \times \frac{2}{3} x^{\frac{-1}{3}} + \frac{5}{3} \times t ^{\frac{2}{3}}\]

which is

\[f'(t) = \frac{4}{\sqrt[3]{t}} + \frac{5t^{\frac{2}{3}}}{3}\]

Answer 4

I guess I did that wrong... So how do I get the second derivative?

Answer 5

or

\[f'(t) = \frac{4}{\sqrt[3]{t}} + \frac{5\sqrt[3]{t^2}}{3}\]

Answer 6

well don't worry about the 2nd derivative at the minute, its used to test for a max or min stationary point or point of inflection.

Answer 7

so you now need to solve for t by letting f'(t) = 0

and the easy solution is t = 0

so thats 1 critical point.

you just need to solve for the other.

so start by cubing every term

\[0 = \frac{64}{t} + \frac{125 t^2}{27}\]

then solve for t

Answer 8

-12/5?

Answer 9

then rewrite the equation

\[-\frac{64}{t} = \frac{125t^2}{27}\]

so

\[t^3 = -\frac{1728}{125}\]

just take the cube root of both sides

Answer 10

thats correct... so the critical points are when

x = -12/5 and x = 0

so not sure if you need the actual ordered pair or just the x values.

Answer 11

Yay! thank you so much! @campbell_st