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Section 3.3 #9: Identify the open intervals where the function is increasing and where the function is decreasing: $$g(x)=x^2-2x-8.$$ Solution: Find the critical points: $$g'(x)=2x-2 \stackrel{\mathrm{set}}{=}0.$$ Solving this equation yields $x=1$. Now make a diagram: From this diagram we may conclude that the function is increasing on the interval $(1,\infty)$ and it is decreasing on the interval $(-\infty,1)$.

Section 3.3 #24: Use the first derivative test to find the relative extrema of $$f(x)=(x+2)^2(x-1).$$ Solution: Find the critical points: $$\begin{array}{ll} f'(x)&=2(x+2)(x-1) + (x+2)^2 \\ &=2(x^2+x-2)+(x^2+4x+4)\\ &=3x^2+6x \stackrel{\mathrm{set}}{=} 0. \end{array}$$ To solve $3x^2+6x=0$, factor out $3x$ to get $3x(x+2)=0$. We get the critical points $x=0,-2$. Now make a diagram: From the diagram, we see there is a relative maximum at $x=-2$ and a relative minimum at $x=0$.

Section 3.4 #4: Determine the open intervals where the function is concave up and where the function is concave down:
$$g(x)=3x^2-x^3.$$ Solution: Find the second derivative and possible inflection points: $$g'(x)=6x-3x^2$$ and $$g''(x)=6-6x \stackrel{\mathrm{set}}{=}0.$$ Find the possible inflection point by solving $6-6x=0$ to get $x=1$. Now make a diagram: From the diagram, we see the graph is concave up on $(-\infty,1)$ and concave down on $(1,\infty)$.

Section 3.4 #17: Find the point(s) of inflection and discuss the concavity of the graph of the function $$f(x)=\dfrac{x^4}{2}+2x^3.$$ Solution: Find the second derivative and possible inflection points: $$f'(x)=2x^3+6x^2,$$ $$f''(x)=6x^2+12x \stackrel{set}{=}0.$$ Solve $6x^2+12x=0$ by factoring out an $x$ to get $x(6x+12)=0$. This has solution $x=0,-2$. Now make a diagram: From the diagram, we see that the graph is concave up on $(-\infty,-3)$ and $(0,\infty)$ and concave down on $(-2,0)$. Also there is an inflection point at $x=-2$ and $x=0$.

Section 3.4 #39: Use the second derivative test to find the relative extrema of
$$f(x)=x + \dfrac{4}{x}.$$ Solution: Find the critical points: $$f'(x) = 1 - \dfrac{4}{x^2} \stackrel{\mathrm{set}}{=} 0.$$ This yields the equation (for $x \neq 0$) $$1 = \dfrac{4}{x^2}.$$ Multiply by $x^2$ to get $$x^2 = 4,$$ and take square root to get the critical points $$x = \pm 2.$$ Find the second derivative: $$f''(x)=\dfrac{8}{x^3}.$$ To use the second derivative test, plug the critical points into $f''$: $$f''(-2) = \dfrac{8}{(-2)^3} = \dfrac{8}{-8} = -1 < 0,$$ from which we can conclude that $f$ has a relative max at $x=-2$, and $$f''(2)=\dfrac{8}{2^3}=\dfrac{8}{8} = 1 > 0,$$ from which we can conclude that $f$ has a relative min at $x=2$.