Exercise 7.7 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.7$
Question 1.
Find intervals of concavity and points of inflexion for the following functions:
(i) $f(x)=x(x-4)^3$
(ii) $f(x)=\sin x+\cos x, 0<x<2 \pi$
(iii) $f(x)=\frac{1}{2}\left(e^x-e^{-x}\right)$
Solution:
(i) $f(x)=x(x-4)^3$
$f(x)=x\left[x^3-12 x^2+48 x-64\right]$
(i.e.,) $f(x)=x^4-12 x^3+48 x^2-64 x$
$f^{\prime}(x)=4 x^3-36 x^2+96 x-64$
$f^{\prime}(x)=12 x^2-72 x+96$
$f^{\prime}(\mathrm{x})=0 \Rightarrow 12\left(\mathrm{x}^2-6 \mathrm{x}+8\right)=0$
$\Rightarrow 12(x-2)(x-4)=0 \Rightarrow x=2$ or 4
Marking the points in the number line

The intervals are $(-\infty, 2),(2,4),(4, \infty)$ when $x \in(-\infty, 2), f^{\prime \prime}(x)=12(-2)(-4)$
$
[\text { say } x=0]=+\mathrm{ve}
$
in the interval $(-\infty, 2)$ the curve concave upwards
when $x \in(2,4), f^{\prime}(x)=(3-2)(3-4)$
$[$ say $\mathrm{x}=3]=-\mathrm{ve}$
$\Rightarrow$ The curve concave upwards in $(2,4)$
when $x \in(4, \infty), f^{\prime}(x)=(5-2)(5-4)$
$[$ say $\mathrm{x}=5]=(+)(+)=+$ ve $\Rightarrow$ The curve concave downwards is $(4, \infty)$ in $(-\infty, 2)$ concave upwards and in $(2,4)$ the concave downwards
$\Rightarrow \mathrm{x}=2$ is a point of inflection at $\mathrm{x}=2, \mathrm{f}(2)=-16$
So $(2,-16)$ is a point of inflection
Again in $(2,4)$ the curve concave downwards and in $(4, \infty)$ the curve concave upwards $\Rightarrow x=4$ is a point of inflection $\mathrm{f}(4)=0$
$\therefore(4,0)$ is a point of inflection
So points of inflection are $(2,-16)$ and $(4,0)$

$
\begin{aligned}
& \text { (ii) } f(\mathrm{x})=\sin \mathrm{x}+\cos \mathrm{x} \\
& f(x)=\cos x-\sin x \\
& f^{\prime \prime}(x)=-\sin x-\cos x \\
& f^{\prime}(x)=0 \Rightarrow-\sin x=\cos x \\
& \Rightarrow \frac{\sin x}{\cos x}=-1 \\
& \text { (i.e.,) } \tan x=-1 \\
& \Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \\
&
\end{aligned}
$
So the intervals are $\left(0, \frac{3 \pi}{4}\right),\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right),\left(\frac{7 \pi}{4}, 2 \pi\right)$ when $x \in\left(0, \frac{3 \pi}{4}\right)$ say $x=\frac{\pi}{2} \quad f^{\prime \prime}(x)=-1-0=-1=-\mathrm{ve}$ $\Rightarrow$ in $\left(0, \frac{3 \pi}{4}\right)$ the curve concave downwards in $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right) f^{\prime \prime}(x)=0-(-1)=1$ say $x=\pi=+\mathrm{ve}$ $\Rightarrow$ in $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$ the curve concave upwards similarly $\left(\frac{7 \pi}{4}, 2 \pi\right)$ the curve concave downwards
. from (1) and (2) $x=\frac{3 \pi}{4}$ is a point of inflection and from (2) and (3) $x=\frac{7 \pi}{4}$ is a point of inflection Now $f\left(\frac{3 \pi}{4}\right)=0$ and $f\left(\frac{7 \pi}{4}\right)=0$
So the curve concave upwards in $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$ and concave downwards in $\left(0, \frac{3 \pi}{4}\right)$, $\left(\frac{7 \pi}{4}, 2 \pi\right)$ and the points of inflection are $\left(\frac{3 \pi}{4}, 0\right),\left(\frac{7 \pi}{4}, 0\right)$

(iii)
$
\begin{aligned}
f(x) & =\frac{1}{2}\left(e^x-e^{-x}\right) \\
f^{\prime}(x) & =\frac{1}{2}\left(e^x+e^{-x}\right) \\
f^{\prime \prime}(x) & =\frac{1}{2}\left(e^x-e^{-x}\right) \\
f^{\prime \prime}(x) & =0 \quad \Rightarrow e^x=e^{-x} \\
& \Rightarrow x=0
\end{aligned}
$
So the intervals are $(-\infty, 0),(0, \infty)$ when $x \in(-\infty, 0) f^{\prime}(x)$ is negative
$\Rightarrow$ the curve concave downwards and when $x \in(0, \infty) f^{\prime}(x)$ is positive
$\Rightarrow$ the curve concave upwards
$\Rightarrow \mathrm{x}=0$ is a point of inflection $\mathrm{f}(0)=\frac{1}{2}(1-1)=0$
So $(0,0)$ is the point of inflection and the curve concave upwards in $(0, \infty)$ and curve concave downwards in $(-\infty$, 0 ) and $(0,0)$ is the point of inflection.

Question 2.
Find the local extrema for the following functions using second derivative test:
(i) $f(x)=-3 x^5+5 x^3$
(ii) $f(x)=x \log x$
(iii) $f(x)=x^2 e^{-2 x}$
Solution:
(i) $f(x)=-3 x^5+5 x^3$
$f^{\prime}(x)=0, f^{\prime}(x)=-v e$ at $x=a$
$\Rightarrow \mathrm{x}=\mathrm{a}$ is a maximum point
$f^{\prime}(x)=0, f^{\prime}(x)=+v e$ at $x=6$
$\Rightarrow \mathrm{x}=\mathrm{b}$ is a minimum point
$f(x)=-3 x^5+5 x^3$
$f^{\prime}(x)=-15 x^4+15 x^2$
$f^{\prime}(x)=-60 x^3+30 x$
$f^{\prime}(x)=0 \Rightarrow-15 x^2\left(x^2-1\right)=0$
$\Rightarrow \mathrm{x}=0,+1,-1$
at $\mathrm{x}=0, \mathrm{f}^{\prime}(\mathrm{x})=0$
at $\mathrm{x}=1, \mathrm{f}^{\prime}(\mathrm{x})=-60+30=-$ ve
at $x=-1, f^{\prime}(x)=60-30=+$ ve
So at $x=1, f^{\prime}(x)=0$ and $f^{\prime}(x)=-v e$
$\Rightarrow \mathrm{x}=1$ is a local maximum point.
and $f(1)=2$
So the local maximum is $(1,2)$
at $x=-1, f^{\prime}(x)=0$ and $f^{\prime}(x)=+$ ve
$\Rightarrow x=-1$ is a local maximum point and $f(-1)=-2$.
So the local minimum point is $(-1,-2)$
$\therefore$ local minimum is $-2$ and local maximum is 2 .

(ii) $f(x)=x \log x$
$
\begin{aligned}
& f^{\prime}(x)=x\left(\frac{1}{x}\right)+\log x=1+\log x \\
& f^{\prime \prime}(x)=\frac{1}{x} \\
& f^{\prime}(x)=0 \Rightarrow 1+\log x=0 \Rightarrow \log x=-1 \\
& \Rightarrow x=e^{-1}=\frac{1}{e} \\
& \text { at } x=\frac{1}{e}, f^{\prime \prime}(x)=\frac{1}{\frac{1}{e}}=e>0(+\mathrm{ve})
\end{aligned}
$
So at $x=\frac{1}{e}, f^{\prime}(x)=0$ and $f^{\prime \prime}(x)=+\mathrm{ve}$
$\Rightarrow x=\frac{1}{e}$ is a local minimum point and $f\left(\frac{1}{e}\right)=\frac{1}{e}(-1)=\frac{-1}{e}$
So local minimum point is $\left(\frac{1}{e}, \frac{-1}{e}\right)$ and local minimum value is $\frac{-1}{e}$

$
\begin{aligned}
& \text { (iii) } f(x)=x^2 e^{-2 x} \\
& f^{\prime}(x)=x^2\left[-2 e^{-2 x}\right]+e^{-2 x}(2 x) \\
& =2 e^{-2 x}\left(x-x^2\right) \\
& f^{\prime}(x)=2 e^{-2 x}(1-2 x)+\left(x--^2\right)\left(-4 e^{-2 x}\right) \\
& =2 e^{-2 x}\left[(1-2 x)+\left(x-x^2\right)(-2)\right] \\
& =2 e^{-2 x}\left[2 x^2-4 x+1\right] \\
& f^{\prime}(x)=0 \Rightarrow 2 e^{-2 x}\left(x-x^2\right)=0 \\
& \Rightarrow x(1-x)=0 \\
& \Rightarrow x=0 \text { or } x=1 \\
& \text { at } x=0, f^{\prime}(x)=2 \times 1[0-0+1]=+v e
\end{aligned}
$
$\Rightarrow \mathrm{x}=0$ is a local minimum point and the minimum value is $\mathrm{f}(0)=0$ at $\mathrm{x}=1$,
$f^{\prime}(x)=2 e^{-2}[2-4+1]=-v e$
$\Rightarrow \mathrm{x}=1$ is a local maximum point and the maximum value is $\mathrm{f}(1)=\frac{1}{e^2}$
Local maxima $\frac{1}{e^2}$ and local minima $=0$

Question 3.
For the function $f(x)=4 x^3+3 x^2-6 x+1$ find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection.
Solution:
$
\begin{aligned}
& 4 x^3+3 x^2-6 x+1 \\
& f^{\prime}(x)=12 x^2+6 x-6 \\
& f^{\prime}(x)=24 x+6 \\
& f^{\prime}(x)=0 \Rightarrow 6\left(2 x^2+x-1\right)=0 \\
& 6(x+1)(2 x-1)=0
\end{aligned}
$
So the intervals are $(-\infty,-1),\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, \infty\right)$
when $x \in(-\infty,-1), f^{\prime}(x)=6(-1)(-5)$
say $x=-2=+\mathrm{ve}$
The function is strictly increasing in $(-\infty,-1)$
when $x \in\left(-1, \frac{1}{2}\right) f^{\prime}(x)=(+1),(-1)$
say $x=0=-\mathrm{ve}$
The function is strictly decreasing in $\left(-1, \frac{1}{2}\right)$
From (1) and (2)
$\mathrm{x}=-1$ is a maximum point,
and $f(-1)=-4+3+6+1=6$
So local maximum is 6
Again when $x \in\left(\frac{1}{2}, \infty\right) \quad f^{\prime}(x)=(2)(1)$
$
\text { say } x=1=+\text { ve }
$
$\Rightarrow$ the function is strictly increasing in $\left(\frac{1}{2}, \infty\right)$
From (2) and (3) $x=\frac{1}{2}$ is a minimum point and $f\left(\frac{1}{2}\right)=\frac{-3}{4}$

So local minimum is $\frac{-3}{4}$
The function is strictly increasing in $(-\infty,-1)$ and $\left(\frac{1}{2}, \infty\right)$ and strictly decreasing in $\left(-1, \frac{1}{2}\right)$. The local maximum is 6 and local minimum is $\frac{-3}{4}$
$
\text { Now } \begin{aligned}
f^{\prime \prime}(x) & =24 x+6 \\
f^{\prime \prime}(x) & =0 \Rightarrow 24 x+6=0 \\
24 x & =-6 \\
\Rightarrow x & =\frac{-1}{4}
\end{aligned}
$
So the intervals are $\left(-\infty, \frac{-1}{4}\right)$ and $\left(\frac{-1}{4}, \infty\right)$
when $x \in\left(-\infty, \frac{-1}{4}\right) f^{\prime \prime}(x)=-24+6$
say $x=-1=-$ ve
$\Rightarrow$ The curve concave downwards in $\left(-\infty, \frac{-1}{4}\right)$
when $x \in\left(\frac{-1}{4}, \infty\right) f^{\prime \prime}(x)=6$
say $x=0=(+$ ve $)$
$\Rightarrow$ The curve concave upwards in $\left(\frac{-1}{4}, \infty\right)$
From (1) and (2) we see that $x=\frac{-1}{4}$ is a point of inflection and $f\left(\frac{-1}{4}\right)=\frac{21}{8}$.
So the curve concave upwards in $\left(\frac{-1}{4}, \infty\right)$ and concave downwards in $\left(-\infty, \frac{-1}{4}\right)$ and the point of inflection is $\left(\frac{-1}{4}, \frac{21}{8}\right)$