115. Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has (i) local maxima (ii) local minima (iii) point of inflexion

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115. Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has (i) local maxima (ii) local minima (iii) point of inflexion

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Vishal Baghel | Contributor-Level 10

4 months ago

We have,

f (x) = (x- 2)⁴ (x + 1)³.

So, f (x) = (x- 2)⁴. 3 (x + 1)² + (x + 1)³. 4 (x- 2)³.

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

= (x- 2)³ (x + 1)² (3x- 6 + 4x + 4)

= (x- 2)³ (x + 1)² (7x- 2).

At f (x) = 0.

(x- 2)³ (x + 1)². (7x- 2) = 0.

x = 2, x = -1 or x = $\frac{2}{7} .$

As (x + 1)²> 0, we shave evaluate for the remaining factor.

At x = 2,

When x< 2, f (x) = ( -ve) (+ ve) (+ ve) = ( -ve) < 0.

When x> 2, f (x) = (+ ve) (+ ve) (+ ve) = (+ ve) > 0.

Øf (x) change from ( -ve) to (+ ve) as x increases

So, x = 2 is a point of local minima

At x = -1.

When x< -1, f (x) = ( -ve) (+ ve) ( -ve) =∉, ve > 0.

When x> -1, f (x) = ( -ve) (+ ve) (+ ve) =∉, ve > 0.

So, f (x) does not change through x -1.

Hence, x = -1 is a point of infixion

At x = $\frac{2}{7,}$

...more

We have,

f (x) = (x- 2)⁴ (x + 1)³.

So, f (x) = (x- 2)⁴. 3 (x + 1)² + (x + 1)³. 4 (x- 2)³.

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

= (x- 2)³ (x + 1)² (3x- 6 + 4x + 4)

= (x- 2)³ (x + 1)² (7x- 2).

At f (x) = 0.

(x- 2)³ (x + 1)². (7x- 2) = 0.

x = 2, x = -1 or x = $\frac{2}{7} .$

As (x + 1)²> 0, we shave evaluate for the remaining factor.

At x = 2,

When x< 2, f (x) = ( -ve) (+ ve) (+ ve) = ( -ve) < 0.

When x> 2, f (x) = (+ ve) (+ ve) (+ ve) = (+ ve) > 0.

Øf (x) change from ( -ve) to (+ ve) as x increases

So, x = 2 is a point of local minima

At x = -1.

When x< -1, f (x) = ( -ve) (+ ve) ( -ve) =∉, ve > 0.

When x> -1, f (x) = ( -ve) (+ ve) (+ ve) =∉, ve > 0.

So, f (x) does not change through x -1.

Hence, x = -1 is a point of infixion

At x = $\frac{2}{7,}$

When x< $\frac{2}{7},$ f (x) = ( -ve) (+ ve) ( -ve) = ( -ve > 0.

When x< $\frac{2}{7},$ f (x) = ( -ve) (+ ve) (+ ve) = ( -ve < 0.

f (x) change from (+ ve) to ( -ve) as x increases through $\frac{2}{7} .$

x = $\frac{2}{7}$ is a point of local maxima

less

We have, f (x) = (x- 2)4 (x + 1)3.So, f (x) = (x- 2)4. 3 (x + 1)2 + (x + 1)3. 4 (x- 2)3.= (x- 2)3 [x + 1)2 [3 (x- 2) + 4 (x + 1)]= (x- 2)3 (x + 1)2 (3x- 6 + 4x + 4)= (x- 2)3 (x + 1)2 (7x- 2).At f (x) = 0. (x- 2)3 (x + 1)2. (7x- 2) = 0.x = 2, x = -1 or x = <math><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>7</mn></mrow></mfrac><mo>.</mo></mrow></math>As (x + 1)2> 0, we shave evaluate for the remaining factor.At x = 2, When x< 2, f (x) = ( -ve) (+ ve) (+ ve) = ( -ve) < 0.When x> 2, f (x) = (+ ve) (+ ve) (+ ve) = (+ ve) > 0.Øf (x) change from ( -ve) to (+ ve) as x increasesSo, x = 2 is a point of local minimaAt x = -1.When x< -1, f (x) = ( -ve) (+ ve) ( -ve) =∉, ve > 0.When x> -1, f (x) = ( -ve) (+ ve) (+ ve) =∉, ve > 0.So, f (x) does not change through x -1.Hence, x = -1 is a point of infixionAt x = <math><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>7</mn><mo>, </mo></mrow></mfrac></mrow></math>When x< <math><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>7</mn></mrow></mfrac><mo>, </mo></mrow></math> f (x) = ( -ve) (+ ve) ( -ve) = ( -ve > 0.When x< <math><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>7</mn></mrow></mfrac><mo>, </mo></mrow></math> f (x) = ( -ve) (+ ve) (+ ve) = ( -ve < 0.f (x) change from (+ ve) to ( -ve) as x increases through <math><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>7</mn></mrow></mfrac><mo>.</mo></mrow></math>x = <math><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>7</mn></mrow></mfrac></mrow></math> is a point of local maxima

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Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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By truth table

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From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

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then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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115. Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has (i) local maxima (ii) local minima (iii) point of inflexion

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