Calculus Solutions EX#4.1

EXERCISE 4.1

10.

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

(b) $f$ is not decreasing anywhere.

(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.

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

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.

$\therefore f$ 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 $[-2,+\infty)$

(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}$