도함수의 뜻

(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.