KCET · Maths · Application of Derivatives
The maximum slope of the curve \(y=-x^{3}+3 x^{2}+2 x-27\) is
- A 1
- B 23
- C 5
- D \(-23\)
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
\(y=-x^{3}+3 x^{2}+2 x-27\)
On differentiating w.r.t. \(x\), we get
\(\frac{d y}{d x}=-3 x^{2}+6 x+2\)
Slope of the curve \(=-3 x^{2}+6 x+2\)
Let \(m=-3 x^{2}+6 x+2\)
\(\therefore \quad \frac{d m}{d x}=-6 x+6\)
Put \(\frac{d m}{d x}=0\) to find the critical points,
\(\begin{aligned}
\Rightarrow &-6 x+6 &=0 \\
\Rightarrow \quad x &=1 \\
&\left(\frac{d^{2} m}{d x^{2}}\right)_{x=1}=-6 < 0
\end{aligned}\)
Slope is maximum at \(x=1\)
The maximum slope,
\(\begin{aligned}
m_{\max } &=-3(1)^{2}+6(1)+2 \\
&=-3+6+2=5
\end{aligned}\)
On differentiating w.r.t. \(x\), we get
\(\frac{d y}{d x}=-3 x^{2}+6 x+2\)
Slope of the curve \(=-3 x^{2}+6 x+2\)
Let \(m=-3 x^{2}+6 x+2\)
\(\therefore \quad \frac{d m}{d x}=-6 x+6\)
Put \(\frac{d m}{d x}=0\) to find the critical points,
\(\begin{aligned}
\Rightarrow &-6 x+6 &=0 \\
\Rightarrow \quad x &=1 \\
&\left(\frac{d^{2} m}{d x^{2}}\right)_{x=1}=-6 < 0
\end{aligned}\)
Slope is maximum at \(x=1\)
The maximum slope,
\(\begin{aligned}
m_{\max } &=-3(1)^{2}+6(1)+2 \\
&=-3+6+2=5
\end{aligned}\)
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