MHT CET · Maths · Application of Derivatives
The range of values of \(x\) for which \(f(x)=x^3+6 x^2-36 x+7\) is increasing in
- A \((-\infty,-6) \cup(2, \infty)\)
- B \((-6,2)\)
- C \((-\infty,-2) \cup(6, \infty)\)
- D \((-6,2]\)
Answer & Solution
Correct Answer
(A) \((-\infty,-6) \cup(2, \infty)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
\mathrm{f}(x) & =x^3+6 x^2-36 x+7 \\
\mathrm{f}^{\prime}(x) & =3 x^2+12 x-36 \\
& =3\left(x^2+4 x-12\right)
\end{aligned}\)
For \(\mathrm{f}(x)\) to be increasing,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x)>0 \\
& \Rightarrow 3\left(x^2+4 x-12\right)>0 \\
& \Rightarrow x^2+4 x-12>0 \\
& \Rightarrow(x+6)(x-2)>0 \\
& \Rightarrow x \in(-\infty,-6) \cup(2, \infty)
\end{aligned}\)
\mathrm{f}(x) & =x^3+6 x^2-36 x+7 \\
\mathrm{f}^{\prime}(x) & =3 x^2+12 x-36 \\
& =3\left(x^2+4 x-12\right)
\end{aligned}\)
For \(\mathrm{f}(x)\) to be increasing,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x)>0 \\
& \Rightarrow 3\left(x^2+4 x-12\right)>0 \\
& \Rightarrow x^2+4 x-12>0 \\
& \Rightarrow(x+6)(x-2)>0 \\
& \Rightarrow x \in(-\infty,-6) \cup(2, \infty)
\end{aligned}\)
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