미분 전체 문제

$f(x)=\frac{2}{3}x^6+5x^4+12x^2+9$ is a polynomial:

$$\begin{aligned} f'(x)&=4x^5+20x^3+24x\\& =4x(x^4+5x^2+6)\\& = 4x((x^2)^2+5(x^2)+6)\\&=4x(x^2+2)(x^2+3) \end{aligned}$$

We can rewrite slightly to make every term into a power of $t$:

$$\begin{aligned} s(t)&=3t^4+5t^3-t^{-1}\\ s'(t)&=4\cdot 3t^{3}+3\cdot 5t^2-(-1)\cdot t^{-2}\\ &=12t^3+15t^2+\frac{1}{t^2} \end{aligned}$$

We could use the product rule here, but it's easier to simplify first. Don't be confused by the role reversal of $x$ and $y$: $x$ is the name of the function, and $y$ is the variable.

$$\begin{aligned} x(y) &= \left(2y+\frac{1}{y}\right)\cdot y^3\\ &=2y^4+y^2\\ x'(y)&=8y^3+2y \end{aligned}$$

We've already seen that $\frac{\mathrm{d}}{\mathrm{d}x}\{\sqrt{x}\}=\frac{1}{2\sqrt{x}}$, but if you forget this formula it is easy to figure out: $\sqrt{x}=x^{1/2}$, so $\frac{\mathrm{d}}{\mathrm{d}x}\{\sqrt{x}\}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$.

Using the quotient rule:

$$\begin{aligned} T(x) &= \dfrac{\sqrt{x}+1}{x^2+3}\\ T'(x)&=\frac{(x^2+3)\left(\frac{1}{2\sqrt{x}}\right)-(\sqrt{x}+1)(2x)}{(x^2+3)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{(x^2+3)\cdot 7 - 2x\cdot (7x+2)}{(x^2+3)^2}=\frac{21 - 4x - 7x^2}{(x^2+3)^2} \end{aligned}$$

Instead of multiplying to get our usual form of this polynomial, we can use the product rule. If $f_1(x)=3x^3+4x^2+x+1$ and $f_2(x)=2x+5$, then $f_1'(x)=9x^2+8x+1$ and $f_2'(x)=2$. Then

$$\begin{aligned} f'(0)&=f_1'(0)f_2(0)+f_1(0)f_2'(0)\\ &=(1)(5)+(1)(2)=7\end{aligned}$$

Using the quotient rule, $$f'(x) = \frac{(x^2+5x)(9x^2)-(3x^3+1)(2x+5)}{(x^2+5x)^2} = \frac{3x^4+30x^3-2x-5}{(x^2+5x)^2}$$

We use quotient rule:

$$\begin{aligned} \frac{(2-x)(6x)-(3x^2+5)(-1)}{(2-x)^2}=\frac{-3x^2+12x+5}{(x-2)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{(3x^2+5)(-2x) - (2-x^2)(6x)}{(3x^2+5)^2}=\frac{-22x}{(3x^2+5)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{6x^2\cdot (x+2)-(2x^3+1)\cdot 1}{(x+2)^2}=\frac{4x^3+12x^2-1}{(x+2)^2} \end{aligned}$$

The derivative of the function is

$$\begin{aligned} \frac{(1-x^2)\cdot\frac{1}{2\sqrt{x}} - \sqrt{x} \cdot (-2x)}{(1-x^2)^2} &= \frac{(1-x^2) - 2x \cdot (-2x)}{2\sqrt{x}(1-x^2)^2} \end{aligned}$$

The derivative is undefined if either $x<0$ or $x = 0,\pm 1$ (since the square-root is undefined for $x<0$ and the denominator is zero when $x=0,1,-1$. Putting this together — the derivative exists for $x>0, x\neq 1$.

Using the product rule seems faster than expanding.

$$\begin{aligned} f'(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3\sqrt[5]{x}+15\sqrt[3]{x}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3{x}^{\frac{1}{5}}+15{x}^{\frac{1}{3}}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\left(\frac{3}{5}{x}^{\frac{-4}{5}}+5{x}^{\frac{-2}{3}}\right)\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\left(6x+8\right) \end{aligned}$$

To avoid the quotient rule, we can divide through the denominator:

$$\begin{aligned} f(x)&=\dfrac{(x^2+5x+1)(\sqrt{x}+\sqrt[3]{x})}{x} =(x^2+5x+1)\dfrac{(\sqrt{x}+\sqrt[3]{x})}{x}\\ &=(x^2+5x+1)(x^{-1/2}+x^{-2/3}) \\ \text{Now, product rule:} \\ f'(x)&=(2x+5)(x^{-1/2}+x^{-2/3})+(x^2+5x+1)\left(\frac{-1}{2}x^{-3/2}-\frac{2}{3}x^{-5/3}\right) \end{aligned}$$

(If you simplified differently, or used the quotient rule, you probably came up with a different-looking answer. There is only one derivative, though, so all correct answers will look the same after sufficient algebraic manipulation.)

We note that $f(0)=f(2\pi)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we conclude that there exists $c$ in $(0,2\pi)$ such that $$f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.$$

We note that $f(0)=f(1)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists $c\in (0,1)$ such that $$f'(c)=\frac{f(1)- f(0)}{1 - 0} = 0.$$

We note that $f(0)=f(2\pi)=\sqrt{3} + \pi^2$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers since $3+\sin x>0$), we get that there exists $c\in (0,2\pi)$ such that $$f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.$$

We note that $f(0)=0$ and $f(\pi/4)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists $c\in (0,\pi/4)$ such that $$f'(c)=\frac{f(\pi/4)- f(0)}{\pi/4 - 0} = 0.$$

By inspection, we see $x=0$ is a root of $f(x)$. The question now is whether there could possibly be other roots. Since $f(x)$ is differentiable over all real numbers, if there is another root $a$, then by Rolle's Theorem, $f'(c)=0$ for some $c$ strictly between $0$ and $a$. However, $f'(x)=3-\cos x$ is never zero. So, there is no second root: $f(x)$ has precisely one root.

The function $f(x)$ is continuous and differentiable over all real numbers. If $a$ and $b$ are distinct roots of $f(x)$, then $f'(c)=0$ for some $c$ strictly between $a$ and $b$ (Rolle's Theorem). So, let's think about $f'(x)$. $$f'(x)=(4x+1)^3+1$$ This is simple enough that we can find its zero explicitly: $$(4x+1)^3 +1= 0 \quad \Leftrightarrow \quad (4x+1)^3 = -1 \quad \Leftrightarrow \quad 4x+1 = -1 \quad \Leftrightarrow \quad 4x=-2$$ Hence $f'(c)=0$ only when $c=\dfrac{-1}{2}$. That the derivative only has a single zero is very useful. It means (via Rolle's theorem) that if $f(x)$ has distinct roots $a$ and $b$ with $a<b$, then we must have $a < \dfrac{-1}{2} < b$. This also means that $f(x)$ cannot have 3 distinct roots $a,b$ and (say) $q$ with $a<b<q$, because then Rolle's theorem would imply that $f'(x)$ would have two zeros — one between $a$ and $b$ and another between $b$ and $q$.

We've learned that $f(x)$ has at most two roots, and we've learned something about where those roots can exist, if there are indeed two of them. But that means $f(x)$ could have 0, 1, or 2 roots.

It's not easy to find a root of $f(x)$ by inspection. But we can get a good enough picture of the graph of $y=f(x)$ to tell exactly how many roots there are, just by exploiting the following properties of $f(x)$ and $f'(x)$.

• As $x$ tends to $\pm \infty$, $f(x)$ tends to $+\infty$.
• The derivative $f'(x)=(4x+1)^3+1$ is negative for $x<-\frac{1}{2}$ and is positive for $x>-\frac{1}{2}$. That is, $f(x)$ is decreasing for $x<-\frac{1}{2}$ and increasing for $x>-\frac{1}{2}$.
• $f\left(-\frac{1}{2}\right)=\frac{1}{16}-\frac{1}{2}<0$.

This means the function must look something like the graph below:

except we are unsure of the locations of the $x$-intercepts. At present we only know that one is to the left of $x=-1/2$ and one is to the right.

Note what happens to $f(x)$ as $x$ increases from strongly negative values to strongly positive values.

• When $x$ is large and negative, $f(x)>0$.
• As $x$ increases, $f(x)$ decreases continuously until $x=-\frac{1}{2}$, where $f(x)<0$. In particular, since $f(-1)=\frac{65}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $-1$. More descriptively put, as $x$ increases from hugely negative numbers to $-\frac{1}{2}$, $f(x)$ passes through zero exactly once.
• As $x$ increases beyond $-\frac{1}{2}$, $f(x)$ increases continuously, starting negative and becoming very large and positive when $x$ becomes large and positive. In particular, since $f(0)=\frac{1}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $0$. So, as $x$ increases from $-\frac{1}{2}$ to near $+\infty$, $f(x)$ again passes through zero exactly once.

So $f(x)$ must have exactly two roots, one with $x<-\frac{1}{2}$ and one with $x>-\frac{1}{2}$.