Answer
(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 $f$.
Work Step by Step
$$f(x)=x^4-2x^2+3$$
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 $f$.
