Calculus Solutions EX #4.2

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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)

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

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

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