Calculus Solutions EX #4.2

Taking too long?

Reload document

Open in new tab

EXERCISE 4.2

3. (a)

$f(x)=3 x^{2}-6 x+1 \quad \quad (1)$

First derivative test

Differentiating (1) w.r.t $x$ .

$f^{\prime}(x)=6 x-6 \quad \quad (2)$

Given that $x_{0}=1.$
Sign analysis of $f^{\prime}(x)$

$\begin{center} \begin{tabular}{|c|c|c|c|}\hline\text {Interval} & $\begin{array}{c}\text { test point } \\ c\end{array}$ & $\begin{array}{l}f^{\prime }(c) \end{array}$ & Sign of $f^{\prime}(x)$ \\\hline$x<1$ & 0 & $\begin{aligned} f^{\prime }(0) & =6(0)-6 \\ & =-6\end{aligned}$ & - \\\hline$x>1$ & 2 & $\begin{aligned} f^{\prime}(2) & =6(2)-6 \\ & =+6\end{aligned}$ & + \\\hline \end{tabular}\end{center}$

By first derivative test, $f$ has relative minimum at $x=1$

Second derivative test

differentiating (2) w.r.t $x$ .

$\begin{aligned} & f^{\prime \prime}(x)=6 \\ & f^{\prime \prime}(1)=6>0 \end{aligned}$

Thus by second derivative test $f$ has relative minimum at $x=1$ .

3. (b)

$f(x)=x^{3}-3 x+1 \quad \quad (1)$

First derivative test

Differentiating (1) w.r.t $x$
Given that $x_{0}=1.$

$f^{\prime}(x)=3 x^{2}-3 \quad \quad (2)$

Sign analysis of $f^{\prime}(x)$

$\begin{center} \begin{tabular}{|c|c|c|c|}\hline\text {Interval} & $\begin{array}{c}\text { test point } \\ c\end{array}$ & $\begin{array}{l}f^{\prime }(c) \end{array}$ & Sign of $f^{\prime}(x)$ \\\hline$x<1$ & 0 & $\begin{aligned} f^{\prime }(0) & =-3 \\ \end{aligned}$ & - \\\hline$x>1$ & 2 & $\begin{aligned} f^{\prime}(2) & =3(2)^2-3 \\ & =+9\end{aligned}$ & + \\\hline \end{tabular}\end{center}$

By the first derivative test we conclude that $f$ has relative minimum at $x=1$

Second derivative test

Differentiating (2) w.r.t $x$

$f^{\prime \prime}(x)=6 x$

Giver what $x=x_{0}=-1.$

$f^{\prime \prime}(-1)=6(-1)=-6<0$

$\therefore \quad f$ has relative maximum at $x=-1$ .

4(a)

$\quad f(x)=\sin^{2} x$

Given that $x=x_{0}=0$

First derivative test

$f^{\prime}(x)=2 \sin x \cos x=\sin 2 x$

Sign analysis of $f^{\prime}(x)$ ( For test value close to o).

$\begin{center} \begin{tabular}{|c|c|c|}\hline $\begin{array}{c}\text { test point } \\ c\end{array}$ & $\begin{array}{l}f^{\prime }(c) \end{array}$ & Sign of $f^{\prime}(x)$ \\\hline -\pi/4 & $\begin{aligned} f^{\prime }(-\pi/4) & =\sin 2(-\pi/4)=-\sin (\pi/2) \\ & =-1 \end{aligned}$ & - \\\hline \pi/4 & $\begin{aligned} f^{\prime}(\pi/4) & =\sin 2(\pi/4)=\sin (\pi/2) \\ & =+1 \end{aligned}$ & + \\\hline \end{tabular}\end{center}$

By fist derivative we conclude that $f$ has relative minimum at $x=0$ .

second derivative test

Differentiating (1) w.r.t. $x$

$f^{\prime \prime}(x)=2 \cos 2 x$

Given what $x=x_{0}=0$

$\begin{aligned} & \text { Given that } x=x_{0}=0 \\ & \therefore f^{\prime \prime}(0)=2 \cos (0)=2>0 \end{aligned}$

$\Rightarrow f$ has relative minimum at $x=0$

4.(b)

$\quad g(x)=\tan ^{2} x$

Given $x=x_{0}=0$

First derivative test

$g^{\prime}(x)=2 \tan x \sec ^{2} x \quad \quad (1)$

Sign analysin of $g^{\prime}(x)$ near to 0 .

$\begin{center} \begin{tabular}{|c|c|c|}\hline $\begin{array}{c}\text { test point } \\ c\end{array}$ & $\begin{array}{l}g^{\prime }(c) \end{array}$ & Sign of $g^{\prime}(x)$ \\\hline -\pi/4 & $\begin{aligned} g^{\prime }(-\pi/4) & =2 \tan(-\pi/4)\sec^{2}(-\pi/4) \\ &=-2(1)(2) =-4 \end{aligned}$ & - \\\hline \pi/4 & $\begin{aligned} g^{\prime}(\pi/4) & =2 \tan(\pi/4)\sec^{2}(\pi/4) \\ &=2(1)(2) =+4 \end{aligned}$ & + \\\hline \end{tabular}\end{center}$

By the first derivative test we come to the conclusion that $g$ has relative minimum at $x=0$

Second derivative test

Differentiating (1) w.r.t. $x$

$\begin{aligned} & g^{\prime \prime}(x)=2 \sec ^{2} x \sec ^{2} x+2 \tan x(2 \sec x \cdot \sec x \tan x) \\ & g^{\prime \prime}(x)=2 \sec ^{4} x+4 \sec ^{2} x \tan ^{2} x \\ & g^{\prime \prime}(x)=2 \sec ^{2} x\left(\sec ^{2} x+2 \tan ^{2} x\right) \end{aligned}$

$\begin{aligned} & g^{\prime \prime}(0)=2 \sec ^{2}(0)\left(\sec ^{2}(0)+2 \tan ^{2}(0)\right) \\ & g^{\prime \prime}(0)=2(1)(1+0)=2>0 \end{aligned}$

$\Rightarrow g(x)$ has relative minimum at $x=0$

(C) Both the functions $f(x)$ and $g(x)$ are squares for $x \neq 0$ and $x$ near to 0 . Both have positive values for values of $x$ close to zero and both are zero at $x=0$ .

5. (a)

$\begin{aligned} & f(x)=(x-1)^{4} \\ \\ & f^{\prime}(x)=4(x-1)^{3} \quad \quad (1) \\ \\ \text {and} \quad g(x)=x^{3}-3 x^{2}+3 x-2 \\ \\ & g^{\prime}(x)=3 x^{2}-6 \quad \quad (2). \end{aligned}$

$x=1$ will be stationary point of $f$ and $g$ if $f^{\prime}(1)=0$ and $g^{\prime}(1)=0$

$\begin{aligned} & f^{\prime}(1)=4(1-1)^{3}=0 \\ \\ g^{\prime}(1)=3(1)^{2}-6(1)+3=3-6+3=0 \end{aligned}$

$\Rightarrow x=1$ is a stationary point of $f$ and $g$ .

(b) Differentiating (1) and (2) in (a)

$\begin{aligned} & f^{\prime \prime}(x)=12(x-1)^{2} \\ & g^{\prime \prime}(x)=6 x-6 \end{aligned}$

$\begin{aligned} & f^{\prime \prime}(1)=12(1-1)=0 \quad \text { no conclusion. } \\ & g^{\prime \prime}(1)=6(1)-6=0 \quad \text { no conclusion. } \end{aligned}$

Leave a Comment Cancel reply