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  • At what point, the slope of the curve y = –x^3+3x^2+9x-27 is maximum? Also find the maximum slope.

At what point, the slope of the curve y = -x^3+3x^2+9x-27 is maximum? Also find the maximum slope.

Answers (1)

Given: y = -x^3+3x^2+9x-27

To find: the point in curve where the slope is maximum and the maximum value of the slope.

Explanation: given y = -x^3+3x^2+9x-27

The slope of the curve can be found by calculating the first derivative of the known curve equation,

Thus, slope of the curve is

\frac{d y}{d x}=\frac{d\left(-x^{3}\right)}{d x}+\frac{d\left(3 x^{2}\right)}{d x}+\frac{d(9 x)}{d x}-\frac{d(27)}{d x}$

Then using the derivative,
\\ \frac{d y}{d x}=-3 x^{2}+3.2 x+9-0$ $\\\frac{d y}{d x}=-3 x^{2}+6 x+9$
To know the critical point, it is necessary to have the value of the slope,

\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(-3 \mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(6 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(9)}{\mathrm{dx}}$
Using the derivative leads to,
\\ \frac{d^{2} y}{d x^{2}}=-3 \cdot 2 x+6+0$\\ $\frac{d^{2} y}{d x^{2}}=-6 x+6$

When the second derivative is equated to 0 it gives the critical point, i.e.,

\\ \frac{d^{2} y}{d x^{2}}=0$\\ $\Rightarrow-6 x+6=0$\\ $\Rightarrow 6 x=6$\\ $\Rightarrow x=1 \ldots \ldots .0(\mathrm{i})$
Then finding the third derivative of the curve,
i.e.,
\frac{d^{3} y}{d x^{3}}=\frac{d(-6 x)}{d x}+\frac{d(6)}{d x}$
Putting the values of the derivative results in

\frac{d^{3} y}{d x^{3}}=-6<0$
As the third derivative is less than 0, so the maximum slope of the given curve is at x=1.
Equating the first derivative with x=1 leads to the maximum value of the slope, i.e.,
\\ \left(\frac{d y}{d x}\right)_{x=1}=-3 x^{2}+6 x+9$ \\$\left(\frac{d y}{d x}\right)_{x=1}=-3(1)^{2}+6(1)+9$ \\$\left(\frac{d y}{d x}\right)_{x=1}=-3+6+9$ \\$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=1}=12$
Therefore, the slope of the curve is maximum at x=1, and the maximum value of the slope is 12 .

Posted by

infoexpert22

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