## Calculus: Early Transcendentals 8th Edition

(a) $f$ increases on $(-1,0)$ and $(1,+\infty)$ and decreases on $(-\infty,-1)$ and $(0,1)$. (b) $(0,3)$ is the local maximum and $(1,2)$ and $(-1,2)$ are the local minimums of $f$. (c) $f$ is concave downward on $(-\frac{\sqrt 3}{3}, \frac{\sqrt 3}{3})$ and concave upward on $(-\infty,-\frac{\sqrt 3}{3})$ and $(\frac{\sqrt 3}{3}, +\infty)$ 2 points $(-\frac{\sqrt 3}{3},\frac{22}{9})$ and $(\frac{\sqrt 3}{3},\frac{22}{9})$ are the inflection points of $f$.
$$f(x)=x^4-2x^2+3$$ First, we need to find $f'(x)$ and $f''(x)$. $$f'(x)=4x^3-4x$$ $$f'(x)=4x(x^2-1)$$ $$f'(x)=4x(x-1)(x+1)$$ $$f''(x)=12x^2-4$$ $$f''(x)=4(3x^2-1)$$ $$f''(x)=4(\sqrt 3 x-1)(\sqrt 3 x+1)$$ (a) Take $f'(x)=0$ $$4x(x-1)(x+1)=0$$ $$x=0\hspace{.5cm}or\hspace{.5cm}x=1\hspace{.5cm}or\hspace{.5cm}x=-1$$ These are the critical points of $f$. To find the trend of increase or decrease of $f$, it's all about the sign of $f'$. A rather useful method here is to build the sign table for $f'$ using the critical points for the intervals as the image below. As we can see, on $(-1,0)$ and $(1,+\infty)$, $f'(x)\gt0$, so $f$ would be increasing there. On $(-\infty,-1)$ and $(0, 1)$, $f'(x)\lt0$, so $f$ would be decreasing on these intervals. (b) To find local maximum and minimum, we would use the First Derivative Test. Looking at the table again, at $x=0$, $f'$ changes from positive to negative. So $f$ has a local maximum at $x=0$. At $x=-1$ and $x=1$, $f'$ changes from negative to positive. So $f$ has local minimums at $x=-1$ and $x=1$. $$f(1)=1^4-2\times1^3+3=1-2+3=2$$ $$f(0)=0^4-2\times0^2+3=0-0+3=3$$ $$f(-1)=(-1)^4-2\times(-1)^2+3=1-2+3=2$$ Therefore, $(0,3)$ is the local maximum and $(1,2)$ and $(-1,2)$ the local minimums of $f$. (c) Take $f''(x)=0$ $$4(\sqrt 3x-1)(\sqrt 3x+1)=0$$ $$x=\frac{\sqrt 3}{3}\hspace{.5cm}or\hspace{.5cm}x=-\frac{\sqrt 3}{3}$$ To find out the concavity, we take a close look at the change of signs of $f''(x)$. A table for the signs of $f''(x)$ is also below. Using the Concavity Test and from the table, we see that - On $(-\frac{\sqrt 3}{3}, \frac{\sqrt 3}{3})$, $f''(x)\lt0$, so $f$ is concave downward. - On $(-\infty,-\frac{\sqrt 3}{3})$ and $(\frac{\sqrt 3}{3}, +\infty)$, $f''(x)\gt0$, so $f$ is concave upward. Finally, at $x=\frac{\sqrt 3}{3}$, $f$ changes from concave downward to upward. At $x=-\frac{\sqrt 3}{3}$, $f$ changes from concave upward to downward. $$f(-\frac{\sqrt 3}{3})=(-\frac{\sqrt 3}{3})^4-2(-\frac{\sqrt 3}{3})^2+3=\frac{1}{9}-\frac{2}{3}+3=\frac{22}{9}$$ $$f(\frac{\sqrt 3}{3})=(\frac{\sqrt 3}{3})^4-2(\frac{\sqrt 3}{3})^2+3=\frac{1}{9}-\frac{2}{3}+3=\frac{22}{9}$$ So, 2 points $(-\frac{\sqrt 3}{3},\frac{22}{9})$ and $(\frac{\sqrt 3}{3},\frac{22}{9})$ are the inflection points of $f$. 