MA005: Calculus I

Practice 1: See Fig. 6.

$\begin{array}{l|c|c|c|l}x & f(x) & f^{\prime}(x) & f^{\prime \prime}(x) & \text { Concavity (up or down) } \\\hline 1 & + & + & - & \text { down } \\2 & + & - & - & \text { down } \\3 & - & - & + & \text { up } \\4 & - & 0 & - & \text { down }\end{array}$

Fig. 6

Practice 2: $f(x)=2 x^{3}-15 x^{2}+24 x-7$.

$\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-30 \mathrm{x}+24$ which is defined for all $\mathrm{x}$.

$\mathrm{f}^{\prime}(\mathrm{x})=0$ if $\mathrm{x}=1,4$ (critical values).

$\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-30$.

$\mathrm{f}^{\prime \prime}(1)=-18$ so $\mathrm{f}$ is concave down at the critical value $\mathrm{x}=1$ so $(1, \mathrm{f}(1))=(1,4)$ is a rel. max.

$\mathrm{f}^{\prime \prime}(4)=+18$ so $\mathrm{f}$ is concave up at the critical value $x=4$ so $(4, f(4))=(4,-23)$ is a rel. min.

Fig. 18 shows the graph of $f$.

Fig. 18

Practice 3: The points labeled $(b)$ and $(g)$ in Fig. 8 are inflection points.

Practice 4:

$\begin{aligned}&f(x)=x^{4}-12 x^{3}+30 x^{2}+5 x-7 . f^{\prime}(x)=4 x^{3}-36 x^{2}+60 x+5 \\&f^{\prime \prime}(x)=12 x^{2}-72 x+60=12\left(x^{2}-6 x+5\right)=12(x-1)(x-5)\end{aligned}$

The only candidates to be Inflection Points are $x=1$ and $x=5$.

If $x < 1$, then $f^{\prime \prime}(x)=12(x-1)(x-5)=12$ (neg)(neg) is positive.
If $1 < x < 5$, then $f^{\prime \prime}(x)=12(x-1)(x-5)=12$ (pos )(neg) is negative.
If $5 < x$, then $f^{\prime \prime}(x)=12(x-1)(x-5)=12$ (pos)(pos) is positive.

$f$ changes concavity at $x=1$ and $x=5$ so $x=1$ and $x=5$ are Inflection Points.

Fig. 19 shows the graph of $f$.

Fig. 19