Question

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i)f(x)=x³,x∈[-2,2] (ii) f(x)=sin x+cos x,x∈[0,π] (iii) f(x)=4x-1/2x²,x∈[-2,\(\frac{9}{2}\)] (iv) f(x)=(x-1)²+3,x∈[-3,1]

Answer 1

(i) The given function is f(x) = x3. f'(x)=3x2 Now, f'(x)=0=x=0 Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [−2, 2]. f(0) = 0 f(−2) = (−2) 3 = −8 f(2) = (2)³ = 8 Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2. Also, the absolute minimum value of f on [−2, 2] is −8 occurring at x = −2. (ii) The given function is f(x) = sin x + cos x. f'(x)=0=sinx=cos x= tanx=1 x=π/4

Then, we evaluate the value of f at critical point x=π/4 and at the endpoints of the interval [0, π].

f(\(\frac{\pi}{4}\))=sin \(\frac{\pi}{4}\)+cos \(\frac{\pi}{4}\)=\(\frac{1}{\sqrt2}\)+\(\frac{1}{\sqrt2}\)=\(\frac{2}{\sqrt2}\)=√2

f(0)=sin0+cos0+1=1

f(π)=sinπ+cosπ=0-1=-1

Hence, we can conclude that the absolute maximum value of f on [0, π] is occurring at x=\(\frac{\pi}{4}\)

and the absolute minimum value of f on [0, π] is −1 occurring at x = π.

(iii) The given function is f'(x)=0=x=4

Then, we evaluate the value of f at critical point x = 4 and at the endpoints of the interval [-2,\(\frac{9}{2}\)]

Hence, we can conclude that the absolute maximum value of f on [-2,\(\frac{9}{2}\)] is -10 occurring at x=-2

(iv) The given function is f(x)=(x-1)²+3

f(x)=2(x-1)

Now,

f'(x)=0=2(x − 1) = 0 ∴ x = 1

Then, we evaluate the value of f at critical point x = 1 and at the endpoints of the interval [−3, 1].

f(1)=(1-1)²+3=0+3=3

f(-3)=(-3-1)²+3=16+3=19

Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x = −3 and the minimum value of f on [−3, 1] is 3 occurring at x = 1.