Question

Need a bit of guidance, the problem is "find the bsolute extreme of f(x)=3x^(2/3)-2x on [-1,2]. Can anyone help?

Answer 1

f'(x) = (2/3)*3(x^(-1/3)) - 2 = 0
(x^(1/3)) = 1
x = 1

look the derivative cant have x=0 so we also have to check this point

i dont know if its minimum or maximum but since we need the absolute:

f(0) = 0
f(-1) = 5
f(2) = 0.7622
f(1) = 1

so (-1,5) (0,0)

Answer 2

Thank you for the help, but I don't quite follow how you got your values for f(x), I got some different numbers, could you show me how you got yours? Thank you again for the help

Answer 3

x=1 from differentiation.. f'(x) = 0
x=0 also from differentiation where the derivative there is discontinuous.. so i check what is going on with 0.

x=-1 x=2 bounds .. function gets minimum or maximum on its bounds..

Answer 4

Okay, I understand this, but I don't follow how you got f(-1) = 5, I got -1, so my bounds would be (if I am thinking correctly) f(1)=1 and f(-1)=-1

Answer 5

(-1)^(2/3) = ((-1)^2)^(1/3) = 1
3*1 - (2)*(-1) = 3 + 2 = 5

Answer 6

3*1 - (2)*(-1) = 3 + 2 = 5, I keep getting 3(-1^2/3) is equal to negative 3, isn't -1^2/3 just equal to -1? thank you again for your help

Answer 7

you have to do (-1)^(2/3)

Answer 8

you can either : ((-1)^2)^(1/3) = (1)^(1/3) = 1
or : ((-1)^(1/3))^2 = (-1)^2 = 1

Answer 9

derivative of \[3 x^{2/3}-2 x \]is\[\frac{2}{\sqrt[3]{x}}-2 \]Combine the terms,set the result to zero and solve for x\[-\frac{2 \left(\sqrt[3]{x}-1\right)}{\sqrt[3]{x}}=0\]\[\sqrt[3]{x}-1=0 \]\[\left(\sqrt[3]{x}\right)^3=1^3 \]\[x=1 \]Plot attached.

Answer 10

robtobey - you suggest that x^(2/3) is not defined for x<0 ?

Answer 11

Made a mistake. 3x^(2/3) - 2x on the plot is defined at x = 0.