미분 전체 문제

A good approximation from the graph is $f(5)=0.5$. We want to find a secant line whose endpoints are both very close to $x=5$, but that also give us clear $y$-values. It looks like $f(5.25) \approx 1$, and $f(4.75)\approx \frac{1}{8}$. The secant line from $x=5$ to $x=5.25$ has approximate slope $\dfrac{f(5.25)-f(5)}{5.25-5}\approx \dfrac{1-.5}{.25}=2$. The secant line from $x=5$ to $x=4.75$ has approximate slope $\dfrac{0.5-\frac{1}{8}}{5-4.75}=\dfrac{3}{2}$.

The graph increases more and more quickly (gets steeper and steeper), so it makes sense that the secant line to the left of $x=5$ has a smaller slope than the secant line to the right of $x=5$. Also, if you're taking secant lines that have endpoints farther out from $x=5$, you'll notice that the slopes of the secant lines change quite dramatically. You have to be very, very close to $x=5$ to get any kind of accuracy.

If we split the difference, we might approximate the slope of the secant line to be the average of $\frac{3}{2}$ and $2$, which is $\frac{7}{4}$.

Another way to try to figure out the tangent line is by carefully drawing it in with a ruler. This is shown here in blue:

It's much easier to take the slope of a line than a curve, and this one looks like it has slope about 1.5. However, we drew this with a computer: by hand it's much harder to draw an accurate tangent line. (That's why we need calculus!)

The actual slope of the tangent line to the function at $x=5$ is about $1.484$. This is extremely hard to figure out just from the graph–by hand, a guess between $1.25$ and $1.75$ would be very accurate.

There is only one tangent line to $f(x)$ at $P$ (shown in blue), but there are infinitely many choices of $Q$ and $R$ (one possibility shown in red). One easy way to sketch the secant line on paper is to draw any line parallel to the tangent line, and choose two intercepts with $y=f(x)$.

Any place the graph looks flat (if you imagine zooming in) is where the tangent line has slope 0. This occurs three times.

Notice that two of the indicated points are at a low point and a high point, respectively. Later, we'll use these places where the tangent line has slope zero to find where a graph achieves its biggest and smallest values.

We already know that $\frac{\mathrm{d}}{\mathrm{d}x}x=1$ and $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$, so we can compute the derivative of $x^3$ by writing $x^3=(x)(x^2)$,

$$\frac{\mathrm{d}}{\mathrm{d}x} x^3 =\frac{\mathrm{d}}{\mathrm{d}x} (x)(x^2) = (1)(x^2)+(x)(2x)=3x^2$$

When this is evaluated at $x=\frac{1}{2}$ we get $\frac{3}{4}$. Since we also compute $\left( \frac{1}{2}\right)^3=\frac{1}{8}$, the equation of the tangent line is

$$\begin{aligned} y - \frac{1}{8} = \frac{3}{4}\cdot \left(x-\frac{1}{2}\right). \end{aligned}$$

Since $x^2+1$ and $x^2+4$ are always positive, $f(x)$ and $h(x)$ are defined over all real numbers. So, $f(x)$ and $h(x)$ correspond to $A(x)$ and $B(x)$. Which is which? $A(0)=1=f(0)$ while $B(0)=2=h(0)$, so $A(x)=f(x)$ and $B(x)=h(x)$.

That leaves $g(x)$ and $k(x)$ matching to $C(x)$ and $D(x)$. The domain of $g(x)$ is all $x$ such that $x^2-1\ge0$. That is, $|x|\ge1$, like $C(x)$. The domain of $k(x)$ is all $x$ such that $x^2-4\ge0$. That is, $|x|\ge2$, like $D(x)$. So, $C(x)=g(x)$ and $D(x)=k(x)$.

(a) Since $f(0)=2$, we solve

$$\begin{aligned} 2&=\sqrt{\log^2(0+p)}\\ &=\sqrt{\log^2 p}\\ &=\left|\log p\right|\\ \log p &= \pm 2\\ p&=e^{\pm 2}\\ p&=e^2 \text{ or } p=\frac{1}{e^2} \end{aligned}$$

We know that $p$ is $e^2$ or $\dfrac{1}{e^2}$, but we have to decide between the two. In both cases, $f(0)=2$. Let's consider the domain of $f(x)$. Since $\log^2(x+p)$ is never negative, the square root does not restrict our domain. However, we can only take the logarithm of positive numbers. Therefore, the domain is

$$\begin{aligned} x \text{ such that } &x+p>0\\ x \text{ such that } &x>-p\\ \end{aligned}$$

If $p=\dfrac{1}{e^2}$, then the domain of $f(x)$ is $\left(-\dfrac{1}{e^2},\infty\right)$. In particular, since $-\dfrac{1}{e^2}>-1$, the domain of $f(x)$ does not include $x=-1$. However, it is clear from the graph that $f(-1)$ exists. So, $p=e^2$.

(b) Now, we need to figure out what $b$ is. Notice that $b$ is the end of the domain of $f(x)$, which we already found to be $(-p,\infty)$. So, $b=-p=-e^2$.

(As a quick check, if we take $e\approx 2.7$, then $-e^2=-7.29$, and this looks about right on the graph.)

(c) The $x$-intercept is the value of $x$ for which $f(x)=0$:

$$\begin{aligned} 0&=\sqrt{\log^2(x+p)}\\ 0&=\log(x+p)\\ 1&=x+p\\ x&=1-p=1-e^2 \end{aligned}$$

The $x$-intercept is $1-e^2$.

(As another quick check, the $x$-intercept we found is a distance of 1 from the vertical asymptote, and this looks about right on the graph.)