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Local Minima

i fx=x4-62 x2...

Question

(i) f(x) = x⁴ $-$ 62x² + 120x + 9

(ii) f(x) = x³ $-$ 6x² + 9x + 15

(iii) f(x) = (x $-$ 1) (x+2)²

(iv) f(x) = 2/x $-$ 2/x² , x>0

(v) f(x) = xe^x

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

(vii) f(x) = (x+1) (x+2)^1/3, $x \frac{>}{} - 2$

(viii) f(x) = $x \sqrt{32 - x^{2}}, - 5 \frac{<}{} x \frac{<}{} 5$

(ix) f(x) = $x^{3} - 2 a x^{2} + a^{2} x, a > 0, x \in R$

(x) f(x) = $x + \frac{a 2}{x}, a > 0,$ x ≠ 0

(xi) f(x) = $x \sqrt{2 - x^{2}} - \sqrt{2} \leq x \leq \sqrt{2}$

(xii) f(x) = $x + \sqrt{1 - x}, x \leq 1$

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Solution

$(i) Given : f (x) = x^{4} - 62 x^{2} + 120 x + 9 \Rightarrow f' (x) = 4 x^{3} - 124 x + 120 For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 4 x^{3} - 124 x + 120 = 0 \Rightarrow x^{3} - 31 x + 30 = 0 \Rightarrow (x - 1) (x^{2} + x - 30) = 0 \Rightarrow (x - 1) (x + 6) (x - 5) = 0 \Rightarrow x = 1, 5 and - 6 Thus, x = 1, x = 5 and x = - 6 are the possible points of local maxima or local minima . Now, f'' (x) = 12 x^{2} - 124 At x = 1 : f'' (1) = 12 {(1)}^{2} - 124 = - 112 < 0 So, x = 1 is the point of local maximum . The local maximum value is given by f (1) = 1^{4} - 62 {(1)}^{2} + 120 \times 1 + 9 = 68 At x = 5 : f'' (5) = 12 {(5)}^{2} - 124 = 176 > 0 So, x = 5 is the point of local minimum . The local minimum value is given by f (5) = 5^{4} - 62 {(5)}^{2} + 120 \times 5 + 9 = - 316 At x = - 6 : f'' (- 6) = 12 {(- 6)}^{2} - 124 = 308 > 0 So, x = - 6 is the point of local maximum . The local minimum value is given by f (- 6) = {(- 6)}^{4} - 62 {(- 6)}^{2} + 120 \times (- 6) + 9 = - 1647$

$(ii) Given : f (x) = x^{3} - 6 x^{2} + 9 x + 15 \Rightarrow f' (x) = 3 x^{2} - 12 x + 9 For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow (x - 1) (x - 3) = 0 \Rightarrow x = 1 and 3 Thus, x = 1 and x = 3 are the possible points of local maxima or local minima . Now, f'' (x) = 6 x - 12 At x = 1 : f'' (1) = 6 (1) - 12 = - 6 < 0 So, x = 1 is the point of local maximum . The local maximum value is given by f (1) = 1^{3} - 6 {(1)}^{2} + 9 \times 1 + 15 = 19 At x = 3 : f'' (3) = 6 (3) - 12 = 6 > 0 So, x = 3 is the point of local minimum . The local minimum value is given by f (3) = 3^{3} - 6 {(3)}^{2} + 9 \times 3 + 15 = 15$

$(iii) Given : f (x) = (x - 1) {(x + 2)}^{2} = (x - 1) (x^{2} + 4 x + 4) = x^{3} + 4 x^{2} + 4 x - x^{2} - 4 x - 4 = x^{3} + 3 x^{2} - 4 \Rightarrow f' (x) = 3 x^{2} + 6 x For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 3 x^{2} + 6 x = 0 \Rightarrow 3 x (x + 2) = 0 \Rightarrow x = 0 and - 2 Thus, x = 0 and x = - 2 are the possible points of local maxima or local minima . Now, f'' (x) = 6 x + 6 At x = 0 : f'' (0) = 6 (0) + 6 = 6 > 0 So, x = 0 is the point of local minimum . The local minimum value is given by f (0) = (0 - 1) {(0 + 2)}^{2} = - 4 At x = - 2 : f'' (- 2) = 6 (- 2) + 6 = - 6 < 0 So, x = - 2 is the point of local maximum . The local maximum value is given by f (- 2) = (- 2 - 1) {(- 2 + 2)}^{2} = 0$

$(iv) Given : f (x) = \frac{2}{x} - \frac{2}{x^{2}} = 2 x^{- 1} - 2 x^{- 2} \Rightarrow f' (x) = - 2 x^{- 2} + 4 x^{- 3} = \frac{4}{x^{3}} - \frac{2}{x^{2}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow \frac{4}{x^{3}} - \frac{2}{x^{2}} = 0 \Rightarrow 4 - 2 x = 0 \Rightarrow x = 2 Thus, x = 2 is the possible point of local maxima or local minima . Now, f'' (x) = \frac{- 12}{x^{4}} + \frac{4}{x^{3}} At x = 2 : f'' (2) = \frac{- 12}{16} + \frac{4}{8} = \frac{- 12 + 8}{16} = \frac{- 1}{4} < 0 So, x = 2 is the point of local maximum . The local maximum value is given by f (2) = \frac{2}{2} - \frac{2}{2^{2}} = 1 - \frac{1}{2} = \frac{1}{2}$

$(v) Given : f (x) = x e^{x} \Rightarrow f' (x) = e^{x} + x e^{x} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow e^{x} + x e^{x} = 0 \Rightarrow e^{x} (1 + x) = 0 \Rightarrow e^{x} \neq 0, x = - 1 \Rightarrow x = - 1 Thus, x = - 1 is the possible point of local maxima or local minima . Now, f'' (x) = e^{x} + e^{x} + x e^{x} At x = - 1 : f'' (- 1) = e^{- 1} + e^{- 1} - e^{- 1} = e^{- 1} > 0 So, x = - 1 is the point of local minimum . The local minimum value is given by f (- 1) = - e^{- 1} = - \frac{1}{e}$

$(vi) Given : f (x) = \frac{x}{2} + \frac{2}{x} \Rightarrow f' (x) = \frac{1}{2} - \frac{2}{x^{2}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow \frac{1}{2} - \frac{2}{x^{2}} = 0 \Rightarrow x^{2} = 4 \Rightarrow x = 2 and - 2 Thus, x = 2 and x = - 2 are the possible points of local maxima or a local minima . Since x > 0, x = 2 Now, f'' (x) = \frac{4}{x^{3}} At x = 2 : f'' (2) = \frac{4}{{(2)}^{3}} = \frac{1}{2} > 0 So, x = 2 is the point of local minimum . The local minimum value is given by f (2) = \frac{x}{2} + \frac{2}{x} = 1 + 1 = 2$

$(vii) Given : f (x) = (x + 1) {(x + 2)}^{\frac{1}{3}} \Rightarrow f' (x) = {(x + 2)}^{\frac{1}{3}} + \frac{1}{3} (x + 1) {(x + 2)}^{\frac{- 2}{3}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow {(x + 2)}^{\frac{1}{3}} + \frac{1}{3} (x + 1) {(x + 2)}^{\frac{- 2}{3}} = 0 \Rightarrow \frac{1}{3} (x + 1) = - {(x + 2)}^{\frac{1}{3}} \times {(x + 2)}^{\frac{2}{3}} \Rightarrow \frac{1}{3} (x + 1) = - (x + 2) \Rightarrow x + 1 = - 3 x - 6 \Rightarrow x = \frac{- 7}{4} Thus, x = \frac{- 7}{4} is the possible point of local maxima or local minima . Now, f'' (\frac{- 7}{4}) = \frac{2}{3} {(x + 2)}^{\frac{- 2}{3}} - \frac{2}{9} (x + 1) {(x + 2)}^{\frac{- 5}{3}} At x = \frac{- 7}{4} : f'' (\frac{- 7}{4}) = \frac{2}{3} {(\frac{- 7}{4} + 2)}^{\frac{- 2}{3}} - \frac{2}{9} (\frac{- 7}{4} + 1) {(\frac{- 7}{4} + 2)}^{\frac{- 5}{3}} = \frac{2}{3} {(\frac{1}{4})}^{\frac{- 2}{3}} + \frac{1}{18} {(\frac{1}{4})}^{\frac{- 5}{2}} > 0 So, x = \frac{- 7}{4} is the point of local minimum . The local minimum value is given by f (\frac{- 7}{4}) = (\frac{- 7}{4} + 1) {(\frac{- 7}{4} + 2)}^{\frac{1}{3}} = \frac{- 3}{4} {(\frac{1}{4})}^{\frac{1}{3}} = \frac{- 3}{4^{\frac{4}{3}}}$

$(viii) Given : f (x) = x \sqrt{32 - x^{2}} \Rightarrow f' (x) = \sqrt{32 - x^{2}} - \frac{x^{2}}{\sqrt{32 - x^{2}}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow \sqrt{32 - x^{2}} - \frac{x^{2}}{\sqrt{32 - x^{2}}} = 0 \Rightarrow \sqrt{32 - x^{2}} = \frac{x}{\sqrt{32 - x^{2}}} \Rightarrow 32 - x^{2} = x^{2} \Rightarrow x^{2} = 16 \Rightarrow x = \pm 4 Thus, x = 4 and x = - 4 are the possible points of local maxima or local minima . Now, f'' (x) = \frac{- x}{\sqrt{32 - x^{2}}} - (\frac{2 x \sqrt{32 - x^{2}} + \frac{x^{3}}{\sqrt{32 - x^{2}}}}{32 - x^{2}}) = \frac{- x}{\sqrt{32 - x^{2}}} - (\frac{2 x (32 - x^{2}) + x^{3}}{(32 - x^{2}) \sqrt{32 - x^{2}}}) At x = 4 : f'' (4) = \frac{- 4}{\sqrt{32 - 4^{2}}} - [\frac{8 (32 - 4^{2}) + 4^{3}}{(32 - 4^{2}) \sqrt{32 - 4^{2}}}] = - 1 - \frac{192}{64} = - 3 < 0 So, x = 4 is the point of local maximum . The local maximum value is given by f (4) = 4 \sqrt{32 - 4^{2}} = 16 At x = - 4 : f'' (- 4) = \frac{4}{\sqrt{32 - 4^{2}}} + [\frac{8 (32 - 4^{2}) - 4^{3}}{(32 - 4^{2}) \sqrt{32 - 4^{2}}}] = 1 + 2 = 3 > 0 So, x = - 4 is the point of local minimum . The local minimum value is given by f (- 4) = - 4 \sqrt{32 - 4^{2}} = - 16$

$(ix) f (x) = x^{3} - 2 a x^{2} + a^{2} x \Rightarrow f' (x) = 3 x^{2} - 4 a x + a^{2} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 3 x^{2} - 4 a x + a^{2} = 0 \Rightarrow 3 x^{2} - 3 a x - a x + a^{2} = 0 \Rightarrow 3 x (x - a) - a (x - a) = 0 \Rightarrow (3 x - a) (x - a) = 0 \Rightarrow x = a and \frac{a}{3} Thus, x = a and x = \frac{a}{3} are the possible points of local maxima or local minima . Now, f'' (x) = 6 x - 4 a At x = a : f'' (a) = 6 (a) - 4 a = 2 a > 0 So, x = a is the point of local minimum . The local minimum value is given by f (a) = a^{3} - 2 a {(a)}^{2} + a^{2} (a) = 0 At x = \frac{a}{3} : f'' (\frac{a}{3}) = 6 (\frac{a}{3}) - 4 a = - 2 a < 0 So, x = \frac{a}{3} is the point of local maximum . The local maximum value is given by f (\frac{a}{3}) = {(\frac{a}{3})}^{3} - 2 a {(\frac{a}{3})}^{2} + a^{2} (\frac{a}{3}) = \frac{a^{3}}{27} - \frac{2 a^{3}}{9} + \frac{a^{3}}{3} = \frac{4 a^{3}}{27}$

$(x) Given : f (x) = x + \frac{a^{2}}{x} \Rightarrow f' (x) = 1 - \frac{a^{2}}{x^{2}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 1 - \frac{a^{2}}{x^{2}} = 0 \Rightarrow x^{2} = a^{2} \Rightarrow x = \pm a Thus, x = a and x = - a are the possible points of local maxima or local minima . Now, f'' (x) = \frac{a^{2}}{x^{3}} At x = a : f'' (a) = \frac{a^{2}}{{(a)}^{3}} = \frac{1}{a} > 0 So, x = a is the point of local minimum . The local minimum value is given by f (a) = x + \frac{a^{2}}{x} = a + a = 2 a At x = - a : f'' (a) = \frac{a^{2}}{{(- a)}^{3}} = - \frac{1}{a} < 0 So, x = - a is the point of local maximum . The local maximum value is given by f (- a) = x + \frac{a^{2}}{x} = - a - a = - 2 a$

$(xi) Given : f (x) = x \sqrt{2 - x^{2}} \Rightarrow f' (x) = \sqrt{2 - x^{2}} - \frac{x^{2}}{\sqrt{2 - x^{2}}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow \sqrt{2 - x^{2}} - \frac{x^{2}}{\sqrt{2 - x^{2}}} = 0 \Rightarrow \sqrt{2 - x^{2}} = \frac{x}{\sqrt{2 - x^{2}}} \Rightarrow 2 - x^{2} = x^{2} \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1 Thus, x = 1 and x = - 1 are the possible points of local maxima or local minima . Now, f'' (x) = \frac{- x}{\sqrt{2 - x^{2}}} - (\frac{2 x \sqrt{2 - x^{2}} + \frac{x^{3}}{\sqrt{2 - x^{2}}}}{2 - x^{2}}) = \frac{- x}{\sqrt{2 - x^{2}}} - (\frac{2 x (2 - x^{2}) + x^{3}}{(2 - x^{2}) \sqrt{2 - x^{2}}}) At x = 1 : f'' (1) = \frac{- 1}{\sqrt{2 - 1^{2}}} - [\frac{2 (2 - 1^{2}) + 1^{3}}{(2 - 1^{2}) \sqrt{2 - 1^{2}}}] = - \frac{1}{2} - \frac{3}{2} = - 2 < 0 So, x = 1 is the point of local maximum . The local maximum value is given by f (4) = 1 \sqrt{2 - 1^{2}} = 1 At x = - 1 : f'' (- 1) = \frac{1}{\sqrt{2 - 1^{2}}} + [\frac{2 (2 - 1^{2}) - 1^{3}}{(2 - 1^{2}) \sqrt{2 - 1^{2}}}] = 1 + 1 = 2 > 0 So, x = - 1 is the point of local minimum . The local minimum value is given by f (- 1) = - 1 \sqrt{2 - 1^{2}} = - 1$

$(xii) Given : f (x) = x + \sqrt{1 - x} \Rightarrow f' (x) = 1 - \frac{1}{2 \sqrt{1 - x}} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 1 - \frac{1}{2 \sqrt{1 - x}} = 0 \Rightarrow \sqrt{1 - x} = \frac{1}{2} \Rightarrow 1 - x = \frac{1}{4} \Rightarrow x = \frac{3}{4} Thus, x = \frac{3}{4} is the possible point of local maxima or local minima . Now, f'' (x) = - \frac{\frac{1}{4 \sqrt{1 - x}}}{4 (1 - x)} At x = \frac{3}{4} : f'' (\frac{3}{4}) = - \frac{\frac{1}{4 \sqrt{1 - \frac{3}{4}}}}{4 (1 - \frac{3}{4})} = - \frac{1}{2} < 0 So, x = \frac{3}{4} is the point of local maximum . The local maximum value is given by f (\frac{3}{4}) = \frac{3}{4} + \sqrt{1 - \frac{3}{4}} = \frac{5}{4}$

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Similar questions

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases :

(i) f(x) = 3x + 1, x = - $\frac{1}{3}$

(ii) f(x) = $x^{2}$ - 1, x = 1, -1

(iii) g(x) = $3 x^{2}$ - 2, x = $\frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}$

(vi) p(x) = $x^{3} - 6 x^{2}$ + 11x - 6, x = 1,2,3

(v) f(x) = 5x - $π, x = \frac{4}{5}$

(vi) f(x) = $x^{2}$ ,x = 0

(vii) f(x) = lx + m, = - $\frac{m}{l}$

(viii) f(x) = 2x + 1, x = $\frac{1}{2}$

f (x) = x^{2}

x < 0, f (x) = x

0 < x < 1, f (x) = \frac{1}{x}

x \geq 1

Q. Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:

(i)

f (x) = 3 x + 1, x = - \frac{1}{3}

f (x) = x^{2} - 1, x = 1, - 1

g (x) = 3 x^{2} - 2, x = \frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}

p (x) = x^{3} - 6 x^{2} + 11 x - 6, x = 1, 2, 3

f (x) = 5 x - π, x = \frac{4}{5}

f (x) = x^{2}, x = 0

f (x) = l x + m, x = - \frac{m}{1}

f (x) = 2 x + 1, x = \frac{1}{2}

$If f (x) = (\begin{matrix} x^{3} + x^{2}, f o r 0 \leq x \leq 2 x + 2, f o r 2 \leq x \leq 4 \end{matrix}$
then the odd extension of f(x) would be -

f (x) = \frac{{(x - 1)}^{3} {(x + 2)}^{4} {(x - 3)}^{5} (x + 6)}{x^{2} {(x - 7)}^{3}}

solve the following inequilaties :
(1)

f (x) > 0

f (x) ⩾ 0

f (x) < 0

f (x) ⩽ 0

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