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Applications of Derivatives (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Applications of Derivatives (p-1)2 Marks
Question
Find the values of $x$, such that $f(x)$ is increasing function:
$f(x)=2 x^3-15 x^2+36 x+1$
✓

Answer

$
\begin{aligned}
& f(x)=2 x^3-15 x^2+36 x+1 \\
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(2 x^3-15 x^2+36 x+1\right) \\
& =2 \times 3 x^2-15 \times 2 x+36 \times 1+0 \\
& =6 x^2-30 x+36 \\
& =6\left(x^2-5 x+6\right)
\end{aligned}
$
$f$ is increasing, if $f^{\prime}(x)>0$
i.e. if $6\left(x^2-5 x+6\right)>0$
i.e. if $x^2-5 x+6>0$
i.e. if $x^2-5 x>-6$
i.e. if $x-5 x+\frac{25}{4}>-6+\frac{25}{4}$
i.e. if $\left(x-\frac{5}{2}\right)^2>\frac{1}{4}$
i.e. if $x-\frac{5}{2}>\frac{1}{2}$ or $x-\frac{5}{2}<-\frac{1}{2}$
i.e. if $x>3$ or $x<2$
i.e. if $x \in(-\infty, 2) \cup(3, \infty)$
$\therefore f$ is increasing, if $x \in(-\infty, 2) \cup(3, \infty)$.

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