Question

If the function is given as y= x^3/3 - 3x^2/2 +2x if the value of x at which the given function has a point of minima is n then find the value of 7n

Answer 1

Answer:

minima at x equal to 2

Maxima at x equal to 1

Answer 2

The value of 7n = 14

Given :

• $\displaystyle \sf{ The \: function \: \: \: y = \frac{ {x}^{3} }{3} - \frac{3 {x}^{2} }{2} + 2x }$

• The value of x at which the given function has a point of minima is n

To find :

The value of 7n

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{y = \frac{ {x}^{3} }{3} - \frac{3 {x}^{2} }{2} + 2x }$

Step 2 of 2 :

Find value of x at which the given function has a point of minima

$\displaystyle \sf{y = \frac{ {x}^{3} }{3} - \frac{3 {x}^{2} }{2} + 2x }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = {x}^{2} - 3x + 2 }$

Again differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{ {d}^{2}y }{d {x}^{2} } = 2x - 3 }$

For the maximum or minimum values of y we have

$\displaystyle \sf{ \frac{dy}{dx} = 0 }$

$\displaystyle \sf{ \implies {x}^{2} - 3x + 2 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} - 2x - x+ 2 = 0 }$

$\displaystyle \sf{ \implies x(x - 2) - 1(x - 2) = 0}$

$\displaystyle \sf{ \implies (x - 1) (x - 2) = 0}$

$\displaystyle \sf{ \implies x = 1 \: , \: 2}$

Now ,

$\displaystyle \sf{ \frac{ {d}^{2}y }{d {x}^{2} } \bigg|_{x = 1} =( 2 \times 1) - 3 = 2 - 3 = - 1 < 0 }$

$\displaystyle \sf{ \frac{ {d}^{2}y }{d {x}^{2} } \bigg|_{x = 2} =( 2 \times 2) - 3 = 4 - 3 = 1 > 0 }$

∴ y has maximum value at x = 1 & y has minimum value at x = 2

Step 3 of 3 :

Find the value of 7n

We observe that y has minimum value at x = 2

Now it is given that the value of x at which the given function has a point of minima is n

∴ n = 2

Thus we get ,

$\displaystyle \sf{7n = 7 \times 2 = 14 }$

Hence the required value of 7n = 14

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