Question

the global maximum of f(x)=1+12|x|-3x^2 on [-1,4] is equal to​

Answer 1

The maximum value of x

3

−3x in the interval [0, 2]

f(x)=x

3

−3x

f

′

(x)=3(x

2

−1)

For maxima or minima,

f

′

(x)=0

⇒x=−1,1

⇒x=1 (since -1 does not belongs in the given interval)

f"(x)=6x

f

′′

(1)=6>0 , so f(x) has a minimum at x=1.

Now we will find the value of function at end points

f(0)=0

f(2)=8−6=2

So, maximum value is 2.