6.1.5 Continued

So let's do an example or two illustrating the function analysis procedure.

Example: analyze the function

$$f(x)=\frac{x^{3}-4x}{x^{2}+3}$$

on the interval $[-10,10]$, including finding the global extrema.

1. The domain is all real numbers, normally, but we've restricted it to $[-10,10]$. So that's our domain.

2. There are no vertical asymptotes, because the function is continuous (in particular, the denominator is never zero). There are no horizontal asymptotes, because the degree of the numerator is strictly greater than the degree of the denominator. However, there are slant asymptotes. If we perform polynomial division, we get that

$$f(x)=x-\frac{7x}{x^{2}+3}.$$

Hence, the fraction part goes to zero as $x\to\infty$, and we're left with $x$. Therefore, the slant asymptote for both infinities is $y=x$.

3. The first derivative is

$$f'(x)=\frac{(x^{2}+3)(3x^{2}-4)-(x^{3}-4x)(2x)}{(x^{2}+3)^{2}}=\frac{3x^{4}+5x^{2}-12-2x^{4}+8x^{2}}{(x^{2}+3)^{2}}=\frac{x^{4}+13x^{2}-12}{(x^{2}+3)^{2}}.$$

Setting this equal to zero is tantamount to setting $x^{4}+13x^{2}-12=0$, as the denominator is never zero. To solve this, we let $z=x^{2}$ and use the quadratic formula: $z^{2}+13z-12=0$ implies that

$$z=\frac{-13\pm\sqrt{169-4(-12)}}{2}=\frac{-13\pm\sqrt{217}}{2}.$$

Now, $217=7\times 31$, so we can't simplify the square root any more. Note that $\sqrt{217}\approx 14.73$, so we have one positive $z$ root, and one negative $z$ root. The negative $z$ root yields complex numbers for $x$, so we discard those, and focus on

$$z=\frac{\sqrt{217}-13}{2} \quad \implies \quad x=\pm\sqrt{\frac{\sqrt{217}-13}{2}}.$$

I agree it's not pretty, but in the real world, answers are hardly ever pretty. Be thankful you have an explicit answer like this at all! So, dividing up our domain according to these numbers, we have the following partition of $[-10,10]:$

$$\left[-10,-\sqrt{\frac{\sqrt{217}-13}{2}}\:\right),\quad \left(-\sqrt{\frac{\sqrt{217}-13}{2}},\sqrt{\frac{\sqrt{217}-13}{2}}\:\right),\quad \left(\sqrt{\frac{\sqrt{217}-13}{2}},10 \right].$$

4. We pick the three values $-5,0,5$ as the points in each non-overlapping region. I get that $f'(-5)=67/56 >0$, so the function is increasing there. Then $f'(0)=-4/3<0$, so the function is decreasing there. Finally, $f'(5)=67/56>0$, so we're increasing again.

5. The second derivative is

$$f''(x)=\frac{((x^{2}+3)^{2})(4x^{3}+26x)-2(x^{4}+13x^{2}-12)(x^{2}+3)(2x)}{((x^{2}+3)^{2})^{2}}

=\frac{(x^{2}+3)(4x^{3}+26x)-4x(x^{4}+13x^{2}-12)}{(x^{2}+3)^{3}}$$

$$=\frac{4x^{5}+26x^{3}+12x^{3}+78x-4x^{5}-52x^{3}+48x}{(x^{2}+3)^{3}}=\frac{-14x^{3}+126x}{(x^{2}+3)^{3}}=-\frac{14x(x^{2}-9)}{(x^{2}+3)^{3}}.$$

The zeros of this function occur at $x=0,\pm 3$. Evaluating the second derivative at the critical points yields that the smaller critical point is a local maximum (the second derivative is negative there), and the larger critical point is a local minimum (the second derivative is positive there). The function is concave up from $-10$ to $-3$, concave down from $-3$ to $0$, concave up from $0$ to $3$, and concave down from $3$ to $10$. Evaluate the second derivative yourself in each of these regions to confirm these statements.

6. You can verify that the third derivative is

$$f'''(x)=\frac{42(x^{4}-18x^{2}+9)}{(x^{2}+3)^{4}}.$$

This is nonzero at $0$ and $\pm 3$. Hence, all three zeros of the second derivative are inflection points.

7. Plotting up the function, you get this picture.

One feature of this function of which I was not aware in advance are the inflection points at $\pm 3$. However, they are necessary in order to approach the slant asymptotes properly. Now, we were tasked with finding the global extrema. We have the critical points and the endpoints. If you evaluate the original function at all those points, you'll find that $-10$ is where the global min occurs, and $10$ is where the global max is.

Summary: we followed the procedure in the previous post, and we have successfully found all local and global extrema, points of inflection, regions of increase or decrease, and regions of concavity of the function. We found one slant asymptote. There is nothing further of interest about this function that we don't already know (for the purposes of the vast majority of applications).

Example: analyze the function

$$f(x)=\frac{e^{x}}{x^{2}-4}$$

on the entire real line.

1. The domain is $\mathbb{R}\setminus\{-2,2\}$. That is, the domain is all real numbers that are not $\pm 2$.

2. There are three asymptotes. There is a horizontal asymptote as $x\to-\infty$, because the numerator is getting small, and the denominator is getting big. It goes to zero very quickly as $x\to-\infty$. In addition, we have two vertical asymptotes at $x=\pm 2$, which are fairly straight-forward to find (just factor the denominator and set it equal to zero to find the vertical asymptotes; since the numerator is never zero, all such points are vertical asymptotes).

3. The first derivative is

$$f'(x)=\frac{(x^{2}-4)(e^{x})-(e^{x})(2x)}{(x^{2}-4)^{2}}=\frac{e^{x}(x^{2}-2x-4)}{(x^{2}-4)^{2}}.$$

Setting it equal to zero, we find that

$$x=\frac{2\pm\sqrt{4-4(-4)}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}.$$

So, the critical points are the vertical asymptotes at $\pm 2$, and the points $1\pm\sqrt{5}$. If we divide the real line up according to all these points, we find that we have

$$(-\infty,-2),\;(-2,1-\sqrt{5}),\;(1-\sqrt{5},2),\;(2,1+\sqrt{5}),\;(1+\sqrt{5},\infty).$$

4. We find, by plugging in numbers, that the function $f$ is increasing on $(-\infty,-2),\;(-2,1-\sqrt{5}),$ and $(1+\sqrt{5},\infty)$. It's decreasing on the other two intervals: $(1-\sqrt{5},2)$ and $(2,1+\sqrt{5})$.

5. We find that the second derivative is

$$f''(x)=\frac{e^{x}(x^{4}-4x^{3}-2x^{2}+16x+24)}{(x^{2}-4)^{3}}.$$

Great. How do we deal with finding the roots of a quartic? There is technically a formula for the roots - it's quite complicated. Let's see if we can be a little clever with this one, though. I claim there are no real roots of this quartic (meaning $g(x)=x^{4}-4x^{3}-2x^{2}+16x+24$). Proving this is tricky. I propose dividing up the real line into three regions: $(-\infty,0],\;(0,3],\;(3,\infty)$. I will outline the proof for you, and you can fill in the gaps. Setting $g(x)=0$ is tantamount to setting $x^{4}+16x+24=4x^{3}+2x^{2}$. Why did I choose that way of splitting it up? Because now everything has positive signs. Call the LHS "the quartic" and the RHS "the cubic". For the first region, $(-\infty,0]$, minimize the quartic. It turns out that the quartic is always greater than $4$, which is always greater than the cubic on that interval, hence there are no intersections. For the region $(3,\infty)$, introduce a shift function, $z=x-3$, or $x=z+3$. If you plug this substitution into both the quartic and the cubic, you will find that all the coefficients of the quartic are greater than all the corresponding coefficients of the cubic, and hence the quartic is always greater than the cubic for $z>0$, which corresponds to $x>3$.

The hardest region is $(0,3]$. I will show you two methods for showing the result in this region. One, you do a tangent line to the quartic at $x=0$. The quartic is concave up for all $x>0$, so it lies above its tangent line. You can show that the cubic is less than the tangent line up to $x=2$. Then piece together another straight line from where the tangent line leaves off at $x=2$, and make the second line go through the point $(3,140)$. You can show that the quartic is greater than both these lines on the desired interval, and that the two lines as pieced together are greater than the cubic on the desired interval.

Another method for the interval $(0,3]$ is to make the LHS look like $(x-1)^{4}$, and separate out the terms that do not show up in the actual quartic. That is, you have the following:

$$(x-1)^{4}=x^{4}-4x^{3}+6x^{2}-4x+1.$$

If you compare this to "the quartic", you find that you need

$$(x-1)^{4}=\underbrace{(x^{4}+16x+24)}_{\text{the quartic}}-\underbrace{(4x^{3}-6x^{2}+20x+23)}_{\text{helper function}}=x^{4}-4x^{3}+6x^{2}-4x+1.$$

So, we want to show that

$$x^{4}+16x+24>\underbrace{4x^{3}-6x^{2}+20x+23}_{\text{helper function}}>4x^{3}+2x^{2}$$

on $(0,3]$. The left-hand inequality is obvious, because $(x-1)^{4}\ge 0$. For the right-hand inequality, you get the comparison of a quadratic to zero, which is straight-forward in the region $(0,3]$.

So there are no roots of the second derivative. This makes the rest of step 5 and all of 6 unnecessary.

7. Here's the plot:

You can see the local max and min that we found, along with the asymptotes, regions of increase and decrease, and regions of upward and downward concavity.