도함수의 뜻

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.

(a) By definition, $$\begin{aligned}g'(x) &= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0} \frac{(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x-h + 3 - x^2 + x - 3}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 - h}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h - 1)}{h} \\ &= \lim_{h \to 0} (2x + h - 1) \\ &= 2x - 1\end{aligned}\,.$$ (b) In the above figure, we see plots of both $g$ and $g'$. We observe that the point $(0.5, 2.75)$ is the vertex of the quadratic function $g$, and at this point the slope of the tangent line to $g(x)$ is zero. This aligns with the point $(0.5, 0)$ where $y=g'(x)$ crosses the $x$-axis. In addition, we note that $g'(x)$ is negative for $x < 0.5$, which corresponds to where $g$ is decreasing and has tangent lines with slopes that are negative. Similarly, the values of $g'$ are positive for $x > 0.5$, which align with the values of slopes we see on the original function $g$. (c) By definition, $$\begin{aligned}p'(x) &= \lim_{h \to 0} \frac{p(x+h)-p(x)}{h} \\ &= \lim_{h \to 0} \frac{5(x+h)^2 - 4(x+h) + 12 - (5x^2 - 4x + 12)}{h} \\ &= \lim_{h \to 0} \frac{5x^2 + 10xh + 5h^2 - 4x-4h + 12 - 5x^2 + 4x - 12}{h} \\ &= \lim_{h \to 0} \frac{10xh + 5h^2 - 4h}{h} \\ &= \lim_{h \to 0} \frac{h(10x + 5h - 4)}{h} \\ &= \lim_{h \to 0} (10x + 5h - 4) \\ &= 10x - 4\end{aligned}\,.$$ (d) For $g(x) = x^2 - x + 3$, we found that $g'(x) = 2x - 1$. For $p(x) = 5x^2 - 4x + 12$, we determined that $p'(x) = 10x - 4$. The constants $3$ and $12$ don't seem to affect the results at all, and that makes sense because those numbers only serve to shift the graphs of $g$ and $p$ vertically, which does nothing to change the slope. The coefficient $-4$ on the linear term in $p(x)$ appears to make the “$-4$” appear in $p'(x)= 10x - 4$. That, too, makes sense in light of the fact that if we considered only the linear function $L(x) = -4x$, the slope would everywhere be $-4$, in contrast to the coefficent $-1$ found on the linear term in $g(x) = x^2 - x + 3$, which leads to the constant $-1$ in $g'(x) = 2x - 1$. Finally, the leading coefficient $5$ in $(x) = 5x^2 - 4x + 12$ leads to the coefficient of “$10$” in $p'(x) = 10x -4$. This makes sense because if we considered only $y = 5x^2$, the $5$ would make the graph $5$ times as steep as the graph of $y = x^2$, and thus it affects the derivative proportionately.