미분 전체 문제

$f\,'(0) = 0$ $f\,'(2) = 0$

$f\,'(\pi/2) = 0$ $f\,'(3\pi/2) = 0$

$f\,'(0) = 0$ $f\,'(3.2) = 0$ $f\,'(4)$ is undefined

$f\,'(2)$ is not defined $f\,'(6) = 0$

Both $f\,'(-1)$ and $f\,'(1)$ are undefined.

$f\,'(0) = 0$

$f\,'(0)$ is not defined

If $Q$ is to the left of the $y$ axis, the line through $Q$ and $P$ is increasing, so the secant line has positive slope. If $Q$ is to the right of the $y$ axis, the line through $Q$ and $P$ is decreasing, so the secant line has negative slope.

The slope of the secant line will be $\dfrac{f(2)-f(-2)}{2-(-2)} = \dfrac{f(2)-f(-2)}{4}$, in every part. So, if two lines have the same slope, that means their differences $f(2)-f(-2)$ will be the same.

The graphs in (a),(c), and (e) all have $f(2)-f(-2)=1$, so they all have the same secant line slope. The graphs in (b) and (f) both have $f(2)-f(-2)=-1$, so they both have the same secant line slope. The graph in (d) has $f(2)-f(-2)=0$, and it is the only graph with this property, so it does not share its secant line slope with any of the other graphs.

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.

The function shown is a line, so it has a constant slope–(a) . Since the function is always increasing, $f'$ is always positive, so also (d) holds. Remark: it does not matter that the function itself is sometimes negative; the slope is always positive because the function is always increasing. Also, since the slope is constant, $f'$ is neither increasing nor decreasing: it is the function that is increasing, not its derivative.

The function is always decreasing, so $f'$ is always negative, option (e). However, the function alternates between being more and less steep, so $f'$ alternates between increasing and decreasing several times, and no other options hold.

Remark: $f$ is always positive, but (d) does not hold!

At the left end of the graph, $f$ is decreasing rapidly, so $f'$ is a strongly negative number. Then as we move towards $x=0$, $f$ decreases less rapidly, so $f'$ is a less strongly negative number. As we pass 0, $f$ increases, so $f'$ is a positive number. As we move to the right, $f$ increases more and more rapidly, so $f'$ is an increasing positive number. This description tells us that $f'$ increases for the entire range shown. So (b) holds, but not (a) or (c). Since $f'$ is negative to the left of the $y$ axis, and positive to the right of it, also (d) and (e) do not hold.

$f'(-1)$ does not exist, because to the left of $x=-1$ the slope is a pretty big positive number (looks like around $+1$) and to the right the slope is $-1/4$. Since the derivative involves a limit, that limit needs to match the limit from the left and the limit from the right. The sharp angle made by the graph at $x=-1$ indicates that the left and right limits do not match, so the derivative does not exist.

$f'(3)$ also does not exist. One way to see this is to notice that the function is discontinuous here. More viscerally, note that $f(3)=1$, so as we take secant lines with one endpoint $(3,1)$, and the other endpoint just to the right of $x=3$, we get slopes that are more and more strongly negative, as shown in the picture below. If we take the limit of the slopes of these secant lines as $x$ goes to $3$ from the right, we get $-\infty$. (This certainly doesn't match the slope from the left, which is $-\frac{1}{4}$.)

At $x=-3$, there is some kind of “change" in the graph; however, it is a smooth curve, so the derivative exists here.

Many answers are possible; here is one.

The key is to realize that the few points you're given suggest a pattern, but don't guarantee it. You only know nine points; anything can happen in between.

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

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

The line $y=mx+b$ is tangent to $y=x^2$ at $x=\alpha$ if $$2\alpha=m\text{ and }\alpha^2=m\alpha+b \iff m=2\alpha\text{ and }b=-\alpha^2$$ The same line $y=mx+b$ is tangent to $y=-x^2+2x-5$ at $x=\beta$ if

$$\begin{aligned} -2\beta+2=m&\text{ and }-\beta^2+2\beta-5=m\beta+b\\ \iff m=2-2\beta&\text{ and }b=-\beta^2+2\beta-5-(2-2\beta)\beta=\beta^2-5 \end{aligned}$$

For the line to be simultaneously tangent to the two parabolas we need $$m=2\alpha=2-2\beta\text{ and }b=-\alpha^2=\beta^2-5$$ Substituting $\alpha=1-\beta$ into $-\alpha^2=\beta^2-5$ gives $-(1-\beta)^2=\beta^2-5$ or $2\beta^2-2\beta-4=0$ or $\beta=-1,2$. The corresponding values of the other parameters are $\alpha=2,-1$, $m=4,-2$ and $b=-4,-1$. The two lines are $y=4x-4$ and $y=-2x-1$.

(a) No such function is possible: Rolle's Theorem guarantees $f'(c)=0$ for at least one point $c \in (1,2)$.

For the other functions, examples are below, but many answers are possible.