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= \(\sqrt\frac{\pi}{2}\).
