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  • Maximum slope of the curve y = -x^3 + 3x^2 + 9x - 27 is: A. 0 B. 12 C. 16 D. 32

Maximum slope of the curve y = -x^3 + 3x^2 + 9x - 27 is:
A. 0
B. 12
C. 16
D. 32

Answers (1)

Given equation of curve is y = -x^3 + 3x^2 + 9x - 27

Apply first derivative and get
\frac{d y}{d x}=\frac{d\left(-x^{3}+3 x^{2}+9 x-27\right)}{d x}$
Apply sum rule and $\varrho$ is the differentiation of the constant term, so
\frac{d y}{d x}=-\frac{d\left(x^{3}\right)}{d x}+3 \frac{d\left(x^{2}\right)}{d x}+9 \frac{d(x)}{d x}-0$
Apply power rule and get
\\\frac{d y}{d x}=-3 x^{2}+3 \cdot(2 x)+9.1$ $\\\\\frac{\mathrm{dy}}{\mathrm{dx}}=-3 \mathrm{x}^{2}+6 \mathrm{x}+9 \ldots \ldots \ldots$(i)

Hence, it is the slope of the curve.

Now to find out the second derivative of the given curve, we will differentiate equation (i) once again

\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(-3 \mathrm{x}^{2}+6 \mathrm{x}+9\right)}{\mathrm{dx}}$
Apply sum rule and 0 is the differentiation of the constant term so
\frac{d^{2} y}{d x^{2}}=-3 \frac{d\left(x^{2}\right)}{d x}+6 \frac{d(x)}{d x}+0$
Apply power rule and get
\\ \frac{d^{2} y}{d x^{2}}=-3(2 x)+6.1$ \\$\frac{d^{2} y}{d x^{2}}=-6 x+6=-6(x-1) \ldots$(ii)

Now we will find the critical point by equating the second derivative to 0, we get

-6(x-1) =0

⇒ x-1=0

⇒ x=1

Now, to find out the third derivative of the given curve, we will differentiate equation (ii) once again

\frac{d^{3} y}{d x^{3}}=\frac{d(-6 x+6)}{d x}$
Apply sum rule and 0 is the differentiation of the constant term, so
\frac{d^{3} y}{d x^{3}}=-6 \frac{d(x)}{d x}+0$
Apply power rule and get
\frac{d^{3} y}{d x^{3}}=-6<0$
Hence, maximum slope is at \mathrm{x}=1$
Now, substitute \mathrm{x}=1$ in (i), and get
\\ \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=-3+6+9=12$

Therefore, 12 is the maximum slope of the curve y = -x^3 + 3x^2 + 9x - 27.

So, the correct answer is option B.

Posted by

infoexpert22

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