# Find the local maxima and minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i) f (x) =x^{2 }(ii) g (x) = x^{3} - 3x

(iii) h(x) = sin x + cos x, 0 < x < π/2

(iv) f (x) = sin x - cos x, 0 < x < 2π

(v) f (x) = x^{3} - 6x^{2} + 9x + 15

(vi) g (x) = x/2 + 2/x, x > 0

(vii) g (x) = 1/(x^{2} + 2)

(viii) f (x) = x √1 - x, 0 < x < 1

**Solution:**

(i) f (x) = x^{2}

Therefore,

f' (x) = 2x

Now,

f' (x) = 0

⇒ x = 0

Thus, x = 0 is the only critical point that could possibly be the point of local maxima or local minima of f.

We have

f" (0) = 2, which is positive.

Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f (0) = 0.

(ii) g (x) = x^{3} - 3x

Therefore,

g' (x) = 3x^{2} - 3

Now,

g' (x) = 0

⇒ 3x^{2} - 3 = 0

⇒ x = ± 1

Also,

g"' (x) = 6x

g" (1) = 6 > 0

g" (- 1) = - 6 < 0

By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is

g (1) = 1^{3} - 3

= 1 - 3

= - 2

However, x = - 1 is a point of local maxima and local maximum value of g at x = - 1 is

g (- 1) = (- 1)^{3} - 3(- 1)

= - 1 + 3

= 2

(iii) h(x) = sin x + cos x, 0 < x < π/2

Therefore,

h' (x) = cos x - sin x

Now,

h' (x) = 0

⇒ cos x - sin x = 0

⇒ sin x = cos x

⇒ tan x = 1

x = π/4 ∈ (0, π/2)

Also,

h" (x) = - sin x -cos x

= - (sin x + cos x)

Hence,

h" (π/4 =) = - (1/√2 + 1/√2)

= -2/√2

= - 1/√2 < 0

Therefore, by second derivative test, x = π / 4 is a point of local maxima and the local maximum value of h at x = π / 4

h (π/4) = sin π/4 + cos π/4

= 1/√2 + 1/√2

= √2

(iv) f (x) = sin x - cos x, 0 < x < 2π

Therefore,

f' (x) = cos x + sin x

Now,

f' (x) = 0

⇒ cos x + sin x = 0

⇒ sin x = - cos x

⇒ tan x = - 1

⇒ x = 3π/4, 7π/4 ∈ (0, 2π)

Also,

f" (x) = - sin x + cos x

Hence,

f" (3π/4) = - sin 3π/4 + cos 3π/4

= (- 1/√2 - 1/√2)

= - √2 < 0

f" (7π/4) = - sin 7π/4 + cos 7π/4

= (1/√2 + 1/√2)

= √2 > 0

Therefore, by second derivative test x = 3π / 4 is a point of local maxima and the local maximum value of f at x = 3π / 4 is

f (3π/4) = sin 3π/4 - cos 3π/4

= (1/√2 + 1/√2)

= √2

However, x = 7π / 4 is a point of local minima and the local minimum value of f at x = 7π / 4 is

f (7π/4) = sin 7π/4 - cos 7π/4

= (- 1/√2 - 1/√2)

= - √2

(v) f (x) = x^{3} - 6x^{2} + 9x + 15

Therefore,

f' (x) = 3x^{2} - 12x + 9

Now,

f' (x) = 0

⇒ 3x^{2} - 12x + 9 = 0

⇒ 3(x - 1)(x - 3) = 0

⇒ x = 1, 3

Also,

f" (x) = 6x - 12

= 6(x - 2)

Hence,

f" (1) = 6(1 - 2) = - 6 < 0

f" (3) = 6(3 - 2) = 6 > 0

Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is

f (1) = 1 - 6 + 9 + 15

= 19

However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is

f (3) = 27 - 54 + 27 + 15

= 15

(vi) g (x) = x/2 + 2/x, x > 0

Therefore,

g' (x) = 1/2 - 2/x^{2}

Now,

g' (x) = 0

⇒ 1/2 - 2/x^{2} = 0

⇒ 2/x^{2} = 1/2

⇒ x^{2} = 4

⇒ x = ± 2

Since, x > 0, we take x = 2

Hence,

g" (x) = 4/x^{3}

g" (2) = 4/(2)^{3} = 1/2 > 0

Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is

g (2) = 2/2 + 2/2

= 1 + 1

= 2

(vii) g (x) = 1/(x^{2} + 2)

Therefore,

g' (x) = (- 2x)/(x^{2} + 2)^{2}

Now,

g' (x) = 0

(- 2x)/(x^{2} + 2)^{2} = 0

Now, for values close to x = 0 and to the left of 0, g' (x) > 0.

Also, for values close to x = 0 and to the right of 0, g' (x) < 0.

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of

g (0) = 1/(0 + 2)

= 1/2

(viii) f (x) = x √1 - x, 0 < x < 1

Therefore,

f' (x) = √1 - x + (1/(2√1 - x)) (- 1)

= √1 - x + x/(2√1 - x)

= (2 (1 - x) - x)/(2√1 - x)

= (2 - 3x)/(2√1 - x)

Now,

f' (x) = 0

= (2 - 3x)/(2√1 - x) = 0

⇒ 2 - 3x = 0

⇒ x = 2/3

Also,

f" (x) = 1/2 [(√1 - x (- 3) - (2 - 3x) (- 1/2 √1 - x)) / (1 - x)]

= [(√1 - x (- 3) - (2 - 3x) (- 1/2√1 - x)) / 2 (1 - x)]

= = [- 6(1- x) + (2 - 3x)] / [4 (1 - x) (√1 - x)]

= (3x - 4)/4 (1 - x)^{3/2}

Hence,

f" (2/3) = [3(2/3) - 4] / [4 (1 - 2/3)^{3/2}]

= (2 - 4) / 4 (1 - 2/3)^{3/2}

= - 1/2 (1/3)^{3/2} < 0

Therefore, by second derivative test, x = 2/3 is a point of local maxima and the local maximum value of f at x = 2/3

f (2/3) = 2/3 √1 - 2/3

= 2/3 √1/3

= 2/3√3

= 2√3/9

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 3

## Find the local maxima and minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i) f (x) =x^{2 }(ii) g (x) = x^{3} - 3x

(iii) h(x) = sin x + cos x, 0 < x < π/2 (iv) f (x) = sin x - cos x, 0 < x < 2π (v) f (x) = x^{3} - 6x^{2} + 9x + 15 (vi) g (x) = x/2 + 2/x, x > 0 (vii) g (x) = 1/(x^{2} + 2) (viii) f (x) = x √1 - x, 0 < x < 1

**Summary:**

i) x = 1 is a point of local minima ii) x = - 1 is a point of local maxima iii) x = π / 4 is a point of local maxima iv) x = 7π / 4 is a point of local minima v) x = 3 is a point of local minima

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