Question

Find the absolute maximum and minimum values of the function x^3-3X^2+1 on the interval [-1/2,4]​

Answer 1

Answer:

Step1

f(x)=x

3

−3x+2;0⩽x⩽2

f

′

(x)=3x

2

−3=0⇒x=±1

Our critical points are +1 & -1, but as given in question we have to find absolute maximum in range [0,2] only. Hence we will reject -1

Step2

Evaluating f(x) at critical points & at end points:

f(1)=1−3+2=0

f(0)=0−0+2=2

f(2)=8−6+2=4

Step3

From above we can see f(x) achieves maximum value of 4 at x=2 in range xϵ[0,2]