### Question 1. Find the absolute maximum and absolute minimum values of the following functions in the given intervals:

### (i) f(x) = 4x – x^{2}/2 in [-2, 9/2]

**Solution:**

Given:f(x) = 4x – x^{2}/2 in [-2, 9/2]On differentiation, we get,

f'(x) = 4 – x

For local maxima and local minima we have f'(x) = 0

4 – x = 0

⇒ x = 4

Let’s evaluate the value of f at the critical point x=4 and at the interval [-2, 9/2]

f(4) = 4(4) – 4

^{2}/2 = 16 – 16/2 = 16 – 8 = 8f(-2) = 4(-2) – (-2)

^{2}/2 = -8 – 4/2 = -8 – 2 = -10f(9/2) = 4(9/2) – (9/2)

^{2}/2 = 18 – 81/8 = 18 – 10.125 = 7.875Hence, the absolute maximum value of f in [-2, 9/2] is 8 at x = 4 and

the absolute minimum value of f in [-2, 9/2] is -10 at x = -2.

### (ii) f(x) = (x – 1)^{2}+3 in [-3, 1]

**Solution:**

Given:f(x) = (x – 1)^{2}+3 in [-3, 1]On differentiation, we get,

f'(x) = 2(x – 1)

For local maxima and local minima we have f'(x) = 0

2(x – 1) = 0

⇒ x = 1

Let’s evaluate the value of f at the critical point x = 1 and at the interval [-3, 1]

f(1) = (1 – 1)

^{2 }+ 3 = 3f(−3) = (-3 – 1)

^{2 }+ 3 = 16 + 3 = 19Hence, the absolute maximum value of f in [-3, 1] is 19 at x = -3 and

the absolute minimum value of f in [-3, 1] is 3 at x = 1.

### (iii) f(x) = 3x^{4 }– 8x^{3 }+ 12x^{2 }– 48x + 25 in [0, 3]

**Solution:**

Given:f(x) = 3x^{4 }– 8x^{3 }+ 12x^{2 }– 48x + 25 in [0,3]On differentiation, we get,

f’(x) = 12x

^{3 }– 24x^{2 }+ 24x – 48= 12(x

^{3}– 2x^{2}+ 2x – 4)= 12(x – 2)(x

^{2 }+ 2)Now, for local minima and local maxima we have f′(x) = 0

x = 2 or x

^{2}+ 2 = 0 having no real rootstherefore we consider only x = 2 ∈ [0, 3]

Let’s evaluate the value of f at the critical point x = 2 and at the interval [0, 3]

f(2) = 3(2)

^{4}– 8(2)^{3}+ 12(2)^{2}– 48(2) + 25= 3(16) – 8(8) + 12(4) – 96 + 25

= 48 – 64 + 48 – 96 + 25

= -39

f(0) = 3(0)

^{4}– 8(0)^{3}+ 12(0)^{2}– 48(0) + 25 = 25f(3) = 3(3)

^{4}– 8(3)^{3}+ 12(3)^{2}– 48(3) + 25= 3(81) – 8(27) + 12(9) – 144 + 25

= 243 – 216 + 108 – 144 + 25

= 16

Hence, the absolute maximum value of f in [0, 3] is 25 at x = 0 and

the absolute minimum value of f in [0, 3] is -39 at x = 2.

### (iv) in [1, 9]

**Solution:**

Given:in [1, 9]On differentiation, we get,

Now, for local minima and local maxima we have f′(x) = 0

=

=

= x = 4/3

Let’s evaluate the value of f at the critical point x = 4/3 and at the interval [1, 9]

Hence, the absolute maximum value of f in [1, 9] is 14√2 at x = 9 and

the absolute minimum value of f in [1, 9] is at x = 4/3

### Question 2. Find the maximum value of 2x^{3} – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

**Solution:**

Let f(x) = 2x

^{3}– 24x + 107∴ f'(x) = 6x

^{2}– 24 = 6(x^{2}– 4)Now, for local maxima and local minima we have f'(x) = 0

⇒ 6(x

^{2}– 4) = 0⇒ x

^{2 }= 4⇒ x = ±2

We first consider the interval [1, 3].

Let’s evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the interval [1, 3].

f(2) = 2(2)

^{3}– 24(2) + 107 = 75f(1) = 2(1)

^{3}– 24(1) + 107 = 85f(3) = 2(3)

^{3}– 24 (3) + 107 = 89Hence, the absolute maximum value of f in the interval [1, 3] is 89 occurring at x = 3,

Next, we consider the interval [– 3, – 1].

Evaluate the value of f at the critical point x = – 2 ∈ [-3, -1]

f(−3) = 2(-3)

^{3}– 24(-3) + 107 = 125f(-2) = 2(-2)

^{3}– 24(-3) + 107 = 139f(-1) = 2(-1)

^{3}– 24(-2) + 107 = 129Hence, the absolute maximum value of f is 139 when x = -2.

### Question 3. Find the absolute maximum and minimum values of the function f given by f(x) = cos^{2}x + sinx, x ∈ [0, π].

**Solution:**

Given:f(x) = cos^{2}x + sinx, x ∈ [0, π]On differentiation, we get,

f′(x) = 2cosx(−sinx) + cosx

= −2sinxcosx + cosx

Now, for local minima and local maxima we have f′(x) = 0

−2sinxcosx + cosx = 0

⇒ cosx(−2sinx + 1) = 0

⇒ sinx = 1/2 or cos x = 0

⇒ x = π/6, π/2 as x ∈ [0, π]

Let’s evaluate the value of f at the critical points x=6/π and x = 2/π and at the interval [0, π]

f(π/6) = cos

^{2}π/6 + sinπ/6 = (√3/2)^{2}+1/2 = 5/4f(π/2) = cos

^{2}π/2 + sinπ/2 = 0 + 1 = 1f(0) = cos

^{2}0 + sin0 = 1 + 0 = 1f(π) = cos

^{2}π + sinπ = (−1)^{2 }+ 0 = 1Hence, the absolute maximum value of f in [0, π] is 5/4 at x = π/6 and

the absolute minimum value of f in [0, π] is 1 at x = 0, π/2, π.

### Question 4. Find the absolute maximum and minimum values of the function f given by f(x) = 12x^{4/3} – 6x^{1/3}, x ∈ [−1, 1].

**Solution:**

Given:f(x) = 12x^{4/3} – 6x^{1/3}, x ∈ [−1, 1].On differentiation, we get,

Now, for local minima and local maxima we have f′(x) = 0

⇒ x = 1/8

Let’s evaluate the value of f at the critical points x = 1/8 and at the interval [-1, 1]

f(1/8) = 12(1/8)

^{4/3}– 6(1/8)^{1/3}= -9/4f(−1) = 12(−1)

^{4/3}– 6(−1)^{1/3 }= 18f(1) = 12(1)

^{4/3}– 6(1)^{1/3 }= 6Hence, the absolute maximum value of f in [-1, 1] is 18 at x = -1 and

the absolute minimum value of f in [-1, 1] is -9/4 at x = 1/8.

### Question 5. Find the absolute maximum and minimum values of the function f given by f(x) = 2x^{3} – 15x^{2}+ 36x + 1 in the interval [1, 5].

**Solution: **

Given:f(x) = 2x^{3}– 15x^{2}+ 36x + 1 in the interval [1, 5]On differentiation, we get,

f′(x) = 6x

^{2 }– 30x + 36 = 6(x^{2}– 5x + 6) = 6(x – 2)(x – 3)Now, for local minima and local maxima we have f′(x) = 0

6(x – 2)(x – 3) = 0

⇒ x = 2 and x = 3

Let’s evaluate the value of f at the critical points x = 2 and x = 3 and at the interval [1, 5]

f(1) = 2(1)

^{3}– 15(1)^{2}+ 36(1) + 1 = 24f(2) = 2(2)

^{3}– 15(2)^{2}+ 36(2) + 1 = 29f(3) = 2(3)

^{3}– 15(3)^{2}+ 36(3) + 1 = 28f(5) = 2(5)

^{3}– 15(5)^{2}+ 36(5) + 1 = 56Hence, the absolute maximum value of f in [1, 5] is 56 at x = 5 and

the absolute minimum value of f in [1, 5] is 24 at x = 1.