Let f be the function defined on [-1,2] by f(x)= 3x^2/3 -2x?

Question

What is the max and min value?

Answer 1

find the second derivative of the equattion, set it to 0 and find the x values..... u should get about 2 values

than, pick random numbers between the values u found and plug it into the first derivative... find if the result is positive or negative......

graph.... form positive (positive slope) to negative (negative slope) is max, from neg to post. is min....

Answer 2

for a quadratic expression in the from,
ax^2+bx+c
min value =(b^2-4ac)/4a
here a=3/3=1
b=-2 c=0
min value=(4-0)/4
=1

Answer 3

dy/dx = 2x^(-1/3) -2

It's a parabola with minimum at (1,-1). Therefore the maxima must be at either x=-1 or x=2.

The maximum is at (-1,3).