Calculus Solutions EX#4.1

EXERCISE 4.1

10. 

(a)  $f$ is increasing on $(-\infty,+\infty)$

(b)  $f$ is not decreasing anywhere.

(C) $f$ is concave up on $(-\infty, 1)$ $\quad$ and $(3,+\infty)$

(d)  $f$ is concave down $(1,3)$

(e) the $x$-coordinates of all inflection points $x=1$  and $x=3 $

15-32. Find (a) the intervals on which $f$ is increasing, (b) open intervals on which $f$ is decreasing (c) the open intervals on which $f$ is concave up (d) the open intervals on which $f$ in concave down, and (e) live $x$-coordinates of all inflection points.

15

$$
 \begin{aligned}
& f(x)=x^{2}-3 x+8 \\
& f^{\prime}(x)=2 x-3, \quad f^{\prime \prime}(x)=2 \\
& \text { Put } \quad f^{\prime}(x)=0 \Rightarrow 2 x-3=0 \quad \Rightarrow 2 x=3 \Rightarrow x=3 / 2
\end{aligned}
$$

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

$$

\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
$x<\frac{3}{2}$ & 0 & $\begin{aligned} f^{\prime }(0) & =2(0)-3 \\ & =-3\end{aligned}$ & – \\
\hline
$x>\frac{3}{2}$ & 2 & $\begin{aligned} f^{\prime}(2) & =2(2)-3 \\ & =+1\end{aligned}$ & + \\
\hline

\end{tabular}
\end{center}

$$

(a) $f$ is increasing on $\left[\frac{3}{2},+\infty\right)$ since $f^{\prime}(x)  < 0$  on $\left(\frac{3}{2},+\infty\right)$

(b) $f$ is decreasing on $\left(-\infty, \frac{3}{2}\right]$ since $f^{\prime}(x)>0$ on $\left(-\infty, \frac{3}{2}\right)$

Note: one can take any value from the interval as a test value.

(c) Since $f^{\prime \prime}(x)=2>0$ on $(-\infty,+\infty)$

$\therefore f$ is concave up on $(-\infty,+\infty)$

(d) $f$ is nowhere concave down

(e) Since concavity does not change at any $x$, therefore $f$ has no inflection point.

16.

$$
\begin{aligned}
& f(x)=5-4 x-x^{2} \\
& f^{\prime}(x)=-4-2 x, \quad f^{\prime \prime}(x)=-2
\end{aligned}
$$

Put $$f^{\prime}(x)=0 \Rightarrow-4-2 x=0 \quad-x=4 \Rightarrow x=-2$$

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
$x<-2$ & -3 & $\begin{aligned} f^{\prime }(-3) & =-4-2(-3) \\ & =-4+6=2\end{aligned}$ & + \\
\hline
$x>-2$ & 0 & $\begin{aligned} f^{\prime}(0) & =-4-2(0) \\ & =-4\end{aligned}$ & – \\
\hline

\end{tabular}
\end{center}

$$

 

(a) $f^{\prime}(x)>0$ on $(-\infty,-2) \Rightarrow f$ is increasing on $(-\infty,-2]$

(b) $f^{\prime}(x)<0$ on $(-2,+\infty) \Rightarrow f $ is decreasing on $[-2,+\infty)$

(c) since $f^{\prime \prime}(x)=-2<0$ on $(-\infty,+\infty)$, therefore $f$ is nowhere concave up.

(d) Since $f^{\prime \prime}(x)=-2<0$ on $(-\infty,+\infty)$, therefore $f$ is Concave down on $(-\infty,+\infty)$

(e) $f$ has no inflection point since $f$ does not change the concavity on $(-\infty,+\infty)$

17/ $$f(x)=(2 x+1)^{3}$$

differentiating twice w.r. t  $x$

$$
\begin{aligned}
& f^{\prime}(x)=3(2 x+1)^{3-1}(2(1)+0) \\
& f^{\prime}(x)=6(2 x+1)^{2} \\
& f^{\prime \prime}(x)=6.2\left(2 x+1)^{2-1}(2(1)+0)\right. \\
& f^{\prime \prime}(x)=24(x+1) \\
& \text { Put } f^{\prime}(x)=0 \Rightarrow 6(2 x+1)^{2}=0 \\
& \Rightarrow \quad 2 x+1=0 \\
& 2 x=-1 \Rightarrow x=-1 / 2
\end{aligned}
$$

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
$x<-\frac{1}{2}$ & -1 & $\begin{aligned} f^{\prime }(-1) & =6(2(-1)+1)^{2} \\ & =+6\end{aligned}$ & +\\
\hline
$x>-\frac{1}{2}$ & 0 & $\begin{aligned} f^{\prime}(0) & =6(2(0)+1)^2 \\ & =+6\end{aligned}$ & + \\
\hline

\end{tabular}
\end{center}

$$

 

(a)

$$
\begin{aligned}
& \text { Since } f^{\prime}(x)>0 \text { on }(-\infty,+\infty) \\
& \Rightarrow f(x) \text { is increasing on }(-\infty, \infty)
\end{aligned}
$$

(b)

$$
f \text { is not decreasing anywhere. }
$$

Now put $f^{\prime \prime}(x)=0$

$$
\Rightarrow 24(2 x+1)=0 \Rightarrow 2 x+1=0 \Rightarrow x=\frac{-1}{2}.
$$

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

$$

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\text {Interval} & $\begin{array}{c}\text { test point } \\ c\end{array}$ & $\begin{array}{l}f^{\prime \prime}(c)= \\ 24(2 c+1)\end{array}$ & Sign of $f^{\prime \prime}(x)$ \\
\hline
$x<-\frac{1}{2}$ & -1 & $\begin{aligned} f^{\prime \prime}(-1) & =24(2(-1)+1) \\ & =-24\end{aligned}$ & – \\
\hline
$x>-\frac{1}{2}$ & 0 & $\begin{aligned} f^{\prime}(0) & =24(2(0)+1) \\ & =+24\end{aligned}$ & + \\
\hline
\end{tabular}
\end{center}

$$

(c) Since $f^{\prime \prime}(x)>0$ for $x> -\frac{1}{2}$ $\Rightarrow f$ is concave up on $(-\frac{1}{2},+\infty)$

(d) Since $f^{\prime \prime}(x)<0$ for $x< -\frac{1}{2}$ $\Rightarrow f$ is concave down on $(-\infty, -\frac{1}{2})$ 

(e) Since concavity chnages at $x=-\frac{1}{2}$, therefore $x=-\frac{1}{2}$ is the $x$ coordinate of point of inflection.