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 [tex]\frac{\pi }{2}[/tex]
x²= [tex]\frac{\pi}{2}[/tex]
x= [tex]\pm \sqrt\frac{\pi}{2}[/tex]
As, the interval given is, 0 ≤ x ≤ 3.
x= [tex] \sqrt\frac{\pi}{2}[/tex]= [tex]\frac{22}{14}[/tex]
f"(x)=12 (-sin x²) × 2 x × x + 12 cos x²= 24 x². (-sin x²)+ 12 cos x²
At, x= [tex] \sqrt\frac{\pi}{2}[/tex]= [tex]\frac{22}{14}[/tex]
f"(x)=-24 × [tex]\frac{\pi}{2}[/tex] ×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( [tex] \sqrt\frac{\pi}{2}[/tex])=6 × sin([tex]\frac{\pi}{2}[/tex] )= 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= [tex] \sqrt\frac{\pi}{2}[/tex].
