미분 전체 문제

${f}^{'}(x)=3{x}^{2}$

$T(x)=4x-5$

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.

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.

(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.

When $x=0$, the curve $y=f(x)$ appears to have a flat tangent line, so the $x=0$ is a critical point. However, it is not a local extremum: it is not true that $f(0) \geq f(x)$ for all $x$ near 0, and it is not true that $f(0) \leq f(x)$ for all $x$ near 0.

To the right of the $x$-axis, there is a spike where the derivative of $f(x)$ does not exist. The $x$-value corresponding to this spike (call it $a$) is a singular point, and $f(x)$ has a local maximum at $x=a$.

The $x$-coordinate corresponding to the blue dot (let's call it $a$) is a critical point, because the tangent line to $f(x)$ at $x=a$ is horizontal. There is no lower point nearby, and actually no lower point on the whole interval shown, so $f(x)$ has both a local minimum and a global minimum at $x=a$.

If a function is not continuous at a point, then it is not differentiable at that point. So, the $x$-coordinate corresponding to the discontinuity (let's call it $b$) is a singular point. Values of $f(x)$ immediately to the right of $b$ are lower, and values immediately to the left of $b$ are higher, so $f(x)$ has no local (or global) extremum at $x=b$.

One possible answer is shown below.

For every $x$ in the red interval shown below, $f(2) \geq f(x)$, so $f(2)$ is a local maximum. However, the point marked with a blue dot shows that $f(x)>f(2)$ for some $x$, so $f(2)$ is not a global maximum.

Two examples are given below, but many are possible.

If $f(x)=-x^2$ or $f(x)=-\sqrt{|x|}$, then $f(x)$ has a global maximum at $x=0$. Since $f(x)$ keeps getting more and more strongly negative as $x$ gets farther and farther from 0, $f(x)$ has no global minimum.

Two examples are given below, but many are possible.

If $f(x)=e^x$, then $f(x) > 0$ for all $x$. As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching 0 but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum. Likewise, $f(x)$ increases more and more as we move right, so there is no maximum.

If $f(x)=\arctan(x)+2$, then $f(x) >\left(-\frac{\pi}{2}\right)+2>0$ for all $x$.

As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching $\left(-\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum.

Likewise, as we move right along the $x$-axis, $f(x)$ gets bigger and bigger, approaching $\left(\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets bigger and bigger as we move right, there is no global maximum.

Since $f(5)$ is a global minimum, $f(5) \leq f(x)$ for all $x$, and so in particular $f(5) \leq f(-5)$.

Similarly, $f(-5) \leq f(x)$ for all $x$, so in particular $f(-5) \leq f(5)$.

Since $f(-5) \leq f(5)$ AND $f(5) \leq f(-5)$, it must be true that $f(-5)=f(5)$.

A sketch of one such graph is below.

Functions $A(x)$ and $B(x)$ share something in common that sets them apart from the others: they have a horizontal tangent line only once. In particular, $A'(-2) \neq 0$ and $B'(2) \neq 0$. The only listed functions that do not have two distinct roots are $l(x)$ and $p(x)$. Since $l(-2) \neq 0$ and $p(2) \neq 0$, we conclude $$A'(x)=l(x) \qquad B'(x)=p(x)$$

Function $C(x)$ is never decreasing. Its tangent line is horizontal when $x = \pm 2$, but the curve never decreases, so $C'(x) \geq 0$ for all $x$ and $C'(2)=C'(-2)=0$. The only function that matches this is $n(x)=(x-2)^2(x+2)^2$. Since its linear terms have even powers, it is never negative, and its roots are precisely $x=\pm 2$. $$C'(x)=n(x)$$

For the functions $D(x)$ and $E(x)$ we consider their behaviour near $x=0$. $D(x)$ is decreasing near $x=0$, so $D'(0)<0$, which matches with $o(0)<0$. Contrastingly, $E(x)$ is increasing near zero, so $E'(0)>0$, which matches with $m(0)>0$. $$D'(x)=o(x)\qquad E'(x)=m(x)$$

This function is symmetric across the $y$-axis, so it is even.

The function is not even, because it is not mirrored across the $y$-axis.

Assuming it continues as shown, the function is periodic, because the unit shown below is repeated:

Additionally, $f(x)$ is odd. In a function with odd symmetry, if we mirror the right-hand portion of the curve (the portion to the right of the $y$-axis) across both the $y$-axis and the $x$-axis, it lines up with the left-hand portion of the curve.

Since reflecting the right-hand portion of the graph across the $y$-axis, then the $x$-axis, gives us $f(x)$, we conclude $f(x)$ is odd.

Since the function is even, we simply reflect the portion shown across the $y$-axis to complete the sketch.

Since the function is odd, to complete the sketch, we reflect the portion shown across the $y$-axis (shown dashed), then the $x$-axis (shown in red).