지수함수와 로그함수의 미분

We simplify the functions to get a better idea of what's going on.

($a$): $y=e^{3\log x}+1=\left(e^{\log x}\right)^3+1=x^3+1$. This is not a line.

($b$): $2y+5=e^{3+\log x}=e^3e^{\log x}=e^3x$. Since $e^3$ is a constant, $2y+5=e^3x$ is a line.

($c$): There isn't a fancy simplification here–this isn't a line. If that isn't a satisfactory answer, we can check: a line is a function with a constant slope. For our function, $y'=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{2x} +4\}=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{2x}\}=\frac{\mathrm{d}}{\mathrm{d}x}\left\{(e^x)^2\right\}=2e^xe^x=2e^{2x}$. Since the derivative isn't constant, the function isn't a line.

($d$): $y=e^{\log x}3^e+\log 2=3^e\,x+\log 2$. Since $3^e$ and $\log 2$ are constants, this is a line.

When we say a function is differentiable without specifying a range, we mean that it is differentiable over its domain. The function $f(x)$ is differentiable when $x \neq 1$ for any values of $a$ and $b$; it is up to us to figure out which constants make it differentiable when $x=1$.

In order to be differentiable, a function must be continuous. The definition of continuity tells us that, for $f$ to be continuous at $x=1$, we need $\displaystyle \lim_{x \to 1}f(x)=f(1)$. From the definition of $f$, we see $f(1)=a+b=\displaystyle \lim_{x \to 1^-}f(x)$, so we need $\displaystyle \lim_{x\to 1^+}f(x)=a+b$. Since $\displaystyle \lim_{x \to 1^+}f(x)=e^1=e$, we specifically need $$e=a+b.$$

Now, let's consider differentiability of $f$ at $x=1$. We need the following limit to exist:

$$\begin{aligned} \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}& \\ \text{In particular, we need the one-sided limits to exist and be equal:} \\ \lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h} \\ \text{If $h<0$, then $1+h<1$, so $f(1+h)=a(1+h)^2+b$. If $h>0$, then $1+h>1$, so $f(1+h)=e^{1+h}$. With this in mind, we begin to evaluate the one-sided limits:} \\ \lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^-}\frac{[a(1+h)^2+b]-[a+b]}{h}\\ &=\lim_{h \to 0^-}\frac{ah^2+2ah}{h}=2a\\ \lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^+}\frac{e^{1+h}-(a+b)}{h} \\ \text{Since we take $a+b$ to be equal to $e$ (to ensure continuity):} \\ &= \lim_{h \to 0^+}\frac{e^{1+h}-e^1}{h}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}\right|_{x=1}=e^1=e \end{aligned}$$

So, we also need $$2a= e$$

Therefore, the values of $a$ and $b$ that make $f$ differentiable are $a=b=\dfrac{e}{2}$.

Let $g(x):=\log(f(x))$. Notice $g'(x)=\frac{f'(x)}{f(x)}$.

In order to show that the two curves have horizontal tangent lines at the same values of $x$, we will show two things: first, that if $f(x)$ has a horizontal tangent line at some value of $x$, then also $g(x)$ has a horizontal tangent line at that value of $x$. Second, we will show that if $g(x)$ has a horizontal tangent line at some value of $x$, then also $f(x)$ has a horizontal tangent line at that value of $x$.

Suppose $f(x)$ has a horizontal tangent line where $x=x_0$ for some point $x_0$. This means $f'(x_0)=0$. Then $g'(x_0)=\frac{f'(x_0)}{f(x_0)}$. Since $f(x_0) \neq 0$, $\frac{f'(x_0)}{f(x_0)}=\frac{0}{f(x_0)}=0$, so $g(x)$ also has a horizontal tangent line when $x=x_0$. This shows that whenever $f$ has a horizontal tangent line, $g$ has one too.

Now suppose $g(x)$ has a horizontal tangent line where $x=x_0$ for some point $x_0$. This means $g'(x_0)=0$. Then $g'(x_0)=\frac{f'(x_0)}{f(x_0)}=0$, so $f'(x_0)$ exists and is equal to zero. Therefore, $f(x)$ also has a horizontal tangent line when $x=x_0$. This shows that whenever $g$ has a horizontal tangent line, $f$ has one too.

Remark: if we were not told that $f(x)$ gives only positive numbers, it would not necessarily be true that $f(x)$ and $\log(f(x))$ have horizontal tangent lines at the same values of $x$. If $f(x)$ had a horizontal tangent line at an $x$-value where $f(x)$ were negative, then $\log(f(x))$ would not exist there, let alone have a horizontal tangent line.