Question

Find the local maxima and minima of the following functions. Also find the local maximum and minimum values.

1. f(x) = sin x + cosx, 0 < x < \(\frac{π}{2}\)
2. f(x) = x³ – 3x
3. f(x) = x³ – 6x² + 9x + 15
4. g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\), x > 0
5. g(x) = \(\frac{1}{x^2 + 2}\)

Answer 1

1. Given;

f(x) = sinx + cosx

⇒ f'(x) = cosx – sinx

For turning point f'(x) = 0

⇒ cosx – sinx = 0

⇒ cosx = sinx

⇒ x = \(\frac{π}{4}\)

f”(x) = -sin x – cosx

⇒ f (\(\frac{π}{4}\)) = -sin\(\frac{π}{4}\) – cos\(\frac{π}{4}\) < 0

Hence f(x) has a local maximum at x = \(\frac{π}{4}\) and local maximum value is

2. Given; 

f(x) = x³ – 3x

⇒ f'(x) = 3x² – 3

For turning point f'(x) = 0

⇒ 3x² – 3 = 0

⇒ x = ±1

f”(x) = 6x

When x = -1

⇒ f”(-1) = -6 < 0

Hence f(x) has a local maximum at x = -1 and local maximum value is

f(-1) = (-1)³ – 3(-1) = -1 + 3 = 2

When x = 1

⇒ f”(1) = 6 > 0

Hence f(x) has a local minimum at x = 1 and local minimum value is

f(1) = (1)³ – 3(1) = 1 – 3 = -2.

3. Given;

f(x) = x³ – 6x² + 9x + 15

⇒ f'(x) = 3x² – 12x + 9

For turning point f'(x) = 0

⇒ 3x² – 12x + 9 = 0 ⇒ 3(x² – 4x + 3) = 0

⇒ 3(x – 1)(x – 3) = 0 ⇒ x = 1, 3

f”(x) = 6x – 12

When x = 1

⇒ f”( 1) = 6 – 12 < 0

Hence f(x) has a local maximum at x = 1 and local maximum value is

f(1) = (1)³ – 6(1)² + 9(1) + 15 = 19

When x = 3

⇒ f”(3) = 6(3) – 12 > 0

Hence f(x) has a local minimum at x = 3 and local minimum value is

f(3) = (3)³ – 6(3)² + 9(3) + 15 = 15.

4. Given; 

g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\)

⇒ g'(x) = 12 – \(\frac{2}{x^2}\)

For turning point g'(x) = 0

Since x > 0, the acceptable value of x = 2

Hence g(x) has a local maximum at x = 2 and local maximum value is g(2) = \(\frac{2}{2} + \frac{2}{2} =2\)

5. Given; 

g(x) = \(\frac{1}{x^2 + 2}\)

For turning point g'(x) = 0

Hence g(x) has a local maximum at x = 2 and maximum value is g(2) = \(\frac{1}{0+2}\) = \(\frac{1}{2}\)