Question

Consider the function f(x) = 6sin(x^2) on the interval 0 ≤ x ≤ 3.

Answer 1

a) F' = 6 sin(x^2) = 0
x^2 = pi
x = sqrt(pi)

b) Fmax = F(1) + integral [1, pi] f(x) dx = 9.7743

Answer 2

Solution:

The given function is

f(x)= 6 sin x²→→→0 ≤ x ≤ 3.

Differentiating once

f'(x)= 6 cos x²× 2 x=12 x cos x²

For Maximum or Minimum

f'(x)=0

12 x cos x²=0

cos x²=0 ∧ x=0

cos x²=cos $\frac{\pi }{2}$

x²=  $\frac{\pi}{2}$

x= $\pm \sqrt\frac{\pi}{2}$

As, the interval given is,  0 ≤ x ≤ 3.

x=  $\sqrt\frac{\pi}{2}$= $\frac{22}{14}$

f"(x)=12 (-sin x²) × 2 x × x + 12 cos x²= 24 x². (-sin x²)+ 12 cos x²

At, x=  $\sqrt\frac{\pi}{2}$= $\frac{22}{14}$

f"(x)=-24 × $\frac{\pi}{2}$ ×1 + 0= A negative number

Showing the function attains maxima at this point.

f(0)=6 sin 0=0

f(3)= 6 sin 3²= 6 sin 9(radian)=6 × 0.40(approx)= 2.40

f( $\sqrt\frac{\pi}{2}$)=6 × sin($\frac{\pi}{2}$ )= 6 ×1=6→→Maximum Value

(b) f(1)=5

I.e at x=1, y=5

Maximum value attained by , f(x)=6 sin x² is 6 at , x=  $\sqrt\frac{\pi}{2}$.