Find the absolute maximum and minimum values of f(x) on the given interval:?

Question

a) f(x)=x^4-2x+3; [-2,3]
b) f(x)=x√(4-x^2); [-1,2]
c) f(x)=x/(x^2+1); [0,2]
d) f(x)=x-2cos(x); [-pi,pi]

Answer 1

(a).

The function is f(x) = x⁴ - 2x + 3 and the interval is [-2, 3].

Differentiate with respect to x.

f'(x) = 4x³ - 2.

To fin the critical numbers, set f'(x) = 0 ⇒ 4x³ - 2 = 0 ⇒ x = (1/2)^1/3.

Left Endpoint: f(-2) = (-2)⁴ - 2(-2) + 3 = 23.

Critical Number: f[(1/2)^1/3] = [(1/2)^1/3]⁴ - 2[(1/2)^1/3] + 3 = 0.39685 - 1.5874 + 3 = 1.8 [ Minimum ].

Right Endpoint: f(3) = (3)⁴ - 2(3) + 3 = 78 [ Maximum ].

Answer 2

(b).

The function is f(x) = x√(4 - x²) and the interval is [-1, 2].

Differentiate with respect to x.

f'(x) = x[(-2x)/2√(4 - x²)] + √(4 - x²)

f'(x) = (- x² + 4 - x²)/√(4 - x²).

f'(x) = -2(x² - 2)/√(4 - x²).

To fin the critical numbers, set f'(x) = 0 and f'(x) does not exist.

f'(x) = 0 ⇒ x² - 2 = 0 ⇒ x = ±√2.

f'(x) does not exist when x = ±2.

The values of x = - 2 and x = - √2 are not lie in the interval [-1, 2], so these are negligible.

Left Endpoint: f(-1) = (-1)√[4 - (-1)²] = - √3 = - 1.732 [ Minimum ].

Critical Number: f(√2) = (√2)√[4 - (√2)²] = √4 = 2 [ Maximum ].

Right Endpoint: f(2) = (2)√[4 - (2)²] = 0.

Answer 3

(c).

The function is f(x) = x/(x²+1) and the interval is [0, 2].

Differentiate with respect to x.

f'(x) = (x²+1 - 2x²)/(x²+1)²

f'(x) = (1 - x²)/(x²+1)²

To fin the critical numbers, set f'(x) = 0 and f'(x) does not exist.

f'(x) = 0 ⇒ 1 - x² = 0 ⇒ x = ±1.

f'(x) does not exist when x²+1 = 0 ⇒ x = ± i.

The imaginary numbers are negligible and x = -1 does not lie in the interval [0, 2], so this is negligible.

Left Endpoint: f(0) = 0/(0²+1) = 0 [ Minimum ].

Critical Number: f(1) = 1/(1²+1) = 0.5 [ Maximum ].

Right Endpoint: f(2) = 2/(2²+1) = 2/5 = 0.4.

Answer 4

(d).

The function is f(x) = x - 2 cos(x) and the interval is [-π, π].

Differentiate with respect to x.

f'(x) = 1 + 2 sin(x)

To fin the critical numbers, set f'(x) = 0 and f'(x) does not exist.

f'(x) = 0 ⇒ 1 + 2 sin(x) ⇒ sin(x) = - 1/2 ⇒ sin(x) = - sin(π/6) ⇒ sin(x) = sin(-π/6).

General solution: θ = nπ + (-1)ⁿα, where n is an integer.

If α = -π/6 ⇒ x = nπ - (-1)ⁿ(π/6).

If n = -1 ⇒ x = -π - (-1)^-1(π/6) = - 5π/6

If n = 0 ⇒ x = -π/6.

The solutions in the interval [-π, π] are x = - 5π/6 and x = - π/6.

Left Endpoint: f(-π) = -π - 2 cos(-π) = -π - 2(-1) = 2 - π = 2 - 3.143 = - 1.143

Critical Number: f(-5π/6) = - 5π/6 - 2 cos(-5π/6) = - 2.619 + 1.732 = - 0.887

Critical Number: f(-π/6) = - π/6 - 2 cos(-π/6) = - 0.524 - 1.732 = - 2.2558 [ Minimum ].

Right Endpoint: f(π) = π - 2 cos(π) = π - 2(-1) = π + 2 = 2 + 3.143 = 5.143 [ Maximum ].