Question

Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = |x + 2| − 1 (ii) g(x) = − |x + 1| + 3 (iii) h(x) = sin(2x) + 5 (iv) f(x) = |sin 4x + 3| (v) h(x) = x + 4, x (−1, 1)

Answer 1

(i) f(x) =|x+2|-1

We know that |x+2|≥0 for every x ∴ R.

Therefore, f(x)= |x+2|-≥-1 for every x ∴ R.

The minimum value of f is attained when. |x+2|=0.

|x+2|=0

=x=-2

∴Minimum value of f = f(−2)= |-2+2|-1=-1

Hence, function f does not have a maximum value

(ii) g(x) =-|x+1|+3

We know that for -x|x+1|≤0 every x ∴ R.

Therefore, g(x)= -|x+1|+3≤3  for every x ∴ R.

The maximum value of g is attained when |x+1|.

|x+1|=0

x=-1

∴Maximum value of g = g(−1) = -|-1+1|+3=3

Hence, function g does not have a minimum value

(iii) h(x) = sin2x + 5 We know that − 1 ≤ sin2x ≤ 1. ∴ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ∴ 4 ≤ sin 2x + 5 ≤ 6 Hence, the maximum and minimum values of h are 6 and 4 respectively. (iv) f(x) = |sin 4x+3|

We know that −1 ≤ sin 4x ≤ 1.

2 ≤ sin 4x + 3 ≤ 4 ∴

2 ≤ ≤|sin 4x+3|≤ 4

Hence, the maximum and minimum values of f are 4 and 2 respectively.

(v) h(x) = x + 1, x ∴ (−1, 1)

Here, if a point x0 is closest to −1, then x₁+1<x₁+\(\frac{1}{2}\)+1  we find for all x0 ∴ (−1, 1). Also, if x1 is closest to 1, then for all x₁ ∴ (−1, 1).

Hence, function h(x) has neither maximum nor minimum value in (−1, 1).