what is the first and second derivative of:
f(x) = (x^(2/3))(x-5)
for the first derivative i got f'(x) = (5/3)(x^(2/3)) - (10/3)(x^(-1/3))...... but i need to be able to get the critical numbers for this and i can't figure out what they are. I know i am suppose to set the equation equal to 0 but i still can't get it. And i have to show my work so I'm not suppose to use a calculator.
The first derivative is correct
Solve for x in :
$\displaystyle f'(x) = (5/3)(x^{2/3}) - (10/3)(x^{-1/3})=0$
$\displaystyle \frac{5}{3} x^{2/3}-\frac{10}{3} \frac{1}{x^{1/3}}=0$
Assuming that x is different from 0 (from the beginning ^^)
Multiply both sides by x^(1/3)
$\displaystyle \frac{5}{3} x^{2/3+1/3}-\frac{10}{3}=0$
$\displaystyle \frac{5}{3} x - \frac{10}{3}=0$
Do you see the critical point ? ^^
yes thank you...
Now i'm having trouble with the 2nd derivative... if i have f'(x) = (5/3)(x^(2/3)) - (10/3)(x^(-1/3))... then does the 2nd derivative f"(x) = (10/9)(x^(-1/3)) + (10/9)(x^(-4/3)) ???