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Local Minima

i fx=x-1x-22i...

Question

(i) f(x) = (x $-$ 1) (x $-$ 2)²

(ii) f(x) = $x \sqrt{1 - x}, x \frac{<}{} 1$

(iii) f(x) = $- (x - 1)^{3} (x + 1)^{2}$

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Solution

$(i) Given : f (x) = (x - 1) {(x - 2)}^{2} = (x - 1) (x^{2} - 4 x + 4) = x^{3} - 4 x^{2} + 4 x - x^{2} + 4 x - 4 = x^{3} - 5 x^{2} + 8 x - 4 \Rightarrow f' (x) = 3 x^{2} - 10 x + 8 For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 3 x^{2} - 10 x + 8 = 0 \Rightarrow 3 x^{2} - 6 x - 4 x + 8 = 0 \Rightarrow (x - 2) (3 x - 4) = 0 \Rightarrow x = 2 and \frac{4}{3} Thus, x = 2 and x = \frac{4}{3} are the possible points of local maxima or local minima . Now, f'' (x) = 6 x - 10 At x = 2 : f'' (2) = 6 (2) - 10 = 2 > 0 So, x = 2 is the point of local minimum . The local minimum value is given by f (2) = (2 - 1) {(2 - 2)}^{2} = 0 At x = \frac{4}{3} : f'' (\frac{4}{3}) = 6 (\frac{4}{3}) - 10 = - 2 < 0 So, x = \frac{4}{3} is the point of local maximum . The local maximum value is given by f (\frac{4}{3}) = (\frac{4}{3} - 1) {(\frac{4}{3} - 2)}^{2} = \frac{1}{3} \times \frac{4}{9} = \frac{4}{27}$

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

$(iii) Given : f (x) = - {(x - 1)}^{3} {(x + 1)}^{2} \Rightarrow f' (x) = - [3 {(x - 1)}^{2} {(x + 1)}^{2} + 2 (x + 1) {(x - 1)}^{3}] For the local maxima or minima, we must have f' (x) = 0 \Rightarrow - 3 {(x - 1)}^{2} {(x + 1)}^{2} - 2 (x + 1) {(x - 1)}^{3} = 0 \Rightarrow {(x - 1)}^{2} (x + 1) [- 3 (x + 1) - 2 (x - 1)] = 0 \Rightarrow {(x - 1)}^{2} (x + 1) [- 3 x - 3 - 2 x + 2] = 0 \Rightarrow {(x - 1)}^{2} (x + 1) [- 5 x - 1] = 0 \Rightarrow x = 1, - 1 and \frac{- 1}{5} Thus, x = 1, x = - 1 and x = \frac{- 1}{5} are the possible points of local maxima or local minima . Now, f'' (x) = - [3 \{2 (x - 1) {(x + 1)}^{2} + 2 (x + 1) {(x - 1)}^{2}\} + 2 \{{(x - 1)}^{3} + 3 {(x - 1)}^{2} (x + 1)\}] = - 6 (x - 1) {(x + 1)}^{2} + 6 (x + 1) {(x - 1)}^{2} - 2 {(x - 1)}^{3} - 6 {(x - 1)}^{2} (x + 1) At x = 1 : f'' (1) = - 6 (1 - 1) {(1 + 1)}^{2} + 6 (1 + 1) {(1 - 1)}^{2} - 2 {(1 - 1)}^{3} - 6 {(1 - 1)}^{2} (1 + 1) = 0 So, it is a point of inflexion . At x = - 1 : f'' (- 1) = - 6 (- 1 - 1) {(- 1 + 1)}^{2} + 6 (- 1 + 1) {(- 1 - 1)}^{2} - 2 {(- 1 - 1)}^{3} - 6 {(- 1 - 1)}^{2} (- 1 + 1) = 16 > 0 So, x = - 1 is the point of local minimum . The local minimum value is given by f (- 1) = - {(1 - 1)}^{3} {(- 1 + 1)}^{2} = 0 At x = - \frac{1}{5} : f'' (- \frac{1}{5}) = - 6 (- \frac{1}{5} - 1) {(- \frac{1}{5} + 1)}^{2} + 6 (- \frac{1}{5} + 1) {(- \frac{1}{5} - 1)}^{2} + 2 {(- \frac{1}{5} - 1)}^{3} - 6 {(- \frac{1}{5} - 1)}^{2} (- \frac{1}{5} + 1) = \frac{576}{125} + \frac{384}{125} - \frac{432}{125} - \frac{864}{125} = \frac{- 336}{125} < 0 So, x = - \frac{1}{5} is the point of local maximum . The local maximum value is given by f (- \frac{1}{5}) = - {(- \frac{1}{5} - 1)}^{3} {(- \frac{1}{5} + 1)}^{2} = - (\frac{- 216}{125}) (\frac{16}{25}) = \frac{3465}{3125}$

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