극한·미분 심화 전체 문제

$4\,\cosh (x)\sinh (x)$

$6\,{\sinh }^{5}(x)\cosh (x)$

First, we restrict the domain of $f$ to force it to be one–to–one. There are many intervals we could choose over which $f$ is one–to–one, but the question asks us to contain $x=0$ and be as large as possible; this leaves us with the following restricted function:

The inverse of a function swaps the role of the input and output; so if the graph of $y=f(x)$ contains the point $(a,b)$, then the graph of $Y=f^{-1}(X)$ contains the point $(b,a)$. That is, the graph of $Y=f^{-1}(X)$ is the graph of $y=f(x)$ with the $x$-coordinates and $y$-coordinates swapped. (So, since $y=f(x)$ crosses the $y$-axis at $y=1$, then $Y=f^{-1}(X)$ crosses the $X$-axis at $X=1$.) This swapping is equivalent to reflecting the curve $y=f(x)$ over the line $y=x$.

Remark: while you're getting accustomed to inverse functions, it is sometimes clearer to consider $y=f(x)$ and $Y=f^{-1}(X)$: using slightly different notations for $x$ (the input of $f$, hence the output of $f^{-1}$) and $X$ (the input of $f^{-1}$, which comes from the output of $f$). However, the convention is to use $x$ for the inputs of both functions, and $y$ as the outputs of both functions, as is written on the graph above.

Since $f(0)$ is closer to $g(0)$ than it is to $h(0)$, you would probably want to estimate $f(0) \approx g(0)=1+2\sin (1)$ if you had the means to efficiently figure out what $\sin(1)$ is, and if you were concerned with accuracy. If you had a calculator, you could use this estimation. Also, later in this chapter we will learn methods of approximating $\sin (1)$ that do not require a calculator, but they do require time.

Without a calculator, or without a lot of time, using $f(0)\approx h(0)=0.7$ probably makes the most sense. It isn't as accurate as $f(0) \approx g(0)$, but you get an estimate very quickly, without worrying about figuring out what $\sin(1)$ is.

Remark: when you're approximating something in real life, there probably won't be an obvious “correct" way to do it. There's usually a trade-off between accuracy and ease.

The linear approximation is a line, passing through $(2,f(2))$, with slope $f'(2)$. That is, the linear approximation to $f(x)$ about $x=2$ is the tangent line to $f(x)$ at $x=2$. It is shown below in red.

We have no idea what $f(0.93)$ is, but 0.93 is pretty close to 1, and we definitely know $f(1)$. The linear approximation of $f(x)$ about $x=1$ is given by

$$\begin{aligned} f(x) &\approx f(1)+f'(1)(x-1) \\ \text{So, we calculate:} \\ f(1)&=\log(1)=0\\ f'(x)&=\frac{1}{x}\\ f'(1)&=\frac{1}{1}=1 \\ \text{Therefore,} \\ f(x) &\approx 0+1(x-1)=x-1 \\ \text{When $x=0.93$:} \\ f(0.93) &\approx 0.93-1=-0.07 \end{aligned}$$

(a) Since the value and slope of the tangent line match the value and slope of the function at the point of tangency, we know $p(3) = L(3) = -1$ and $p'(3) = L'(3) = -2$. (b) Using the local linearization, $p(2.79) \approx L(2.79) = -2(2.79) + 5 = -0.58$. (c) Since $p''(x) < 0$ for $x < 3$, we know that $p$ is concave down for $x < 3$. This means to the left of the point of tangency at $x = 3$, the tangent line lies above the curve, and thus the estimate in (b) is too large. (d) In the following figure, we see a possible function $p$ that matches the value and slope of the tangent line at the point $(3,-1)$ and that is concave down to the left of $x=3$ and concave up to the right.

(a) The central difference approximation with $h=10$ gives $$F'(30) \approx \frac{F(40)-F(20)}{20} = \frac{182.4-141.1}{20} = 2.065$$ degrees per minute. (b) The local linearization at $a=30$ is $L(t) = F'(30)(t-30)+F(30) = L(t) \approx 2.065(t-30) + 167.6$. (c) Using the linearization from (b), $F(33) \approx L(33) \approx 2.065(3) + 167.6 = 173.795$ degrees F. (d) Since the values of $F(t)$ given in the table appear to be increasing at a decreasing rate near $t = 30$, it follows that the graph of $F$ is concave down near $t = 30$. This implies that the graph of $L(t)$ (the tangent line to $F$ at $t = 30)$ lies above the graph of $F$. Therefore, the approximation given in part (c) is probably an overestimate.

See the figure below for possible graphs of $f$ and $g$. To evaluate the limit, we can apply L'Hôpital's Rule twice. Since $f(3) = g(3) = 0$, we know by the rule that

$$\lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{f'(x)}{g'(x)}\,.$$ Since $f'(3) = g'(3) = 0$, the limit is still indeterminate, and we can apply the rule again to get

$$\lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{f''(x)}{g''(x)} = \frac{f''(3)}{g''(3)} = \frac{-2}{1}\,.$$ Hence, $\lim_{x \to 3} h(x) = -2$.

First, we note that since both the numerator and denominator of $R$ are quadratic, the limit as $x \to \infty$ can be found by taking the quotient of the leading coefficients. In particular,

$$\lim_{x \to \infty} R(x) = \frac{3}{5}$$

and thus the horizontal asymptote of $R$ is $y = \frac{3}{5}$.

Next, since $b$ and $c$ are distinct, we see that the numerator of $R$ is zero (and the denominator is not) when $x = b$, and thus $R(b) = 0$. In addition, the denominator is zero (and the numerator is not) when $x = c$, and this results in there being a vertical asymptote at $x = c$.

Finally, we observe that $x = a$ makes both the numerator and denominator zero. Since $R(a)$ is thus not defined, the graph of $R$ has either a hole or a vertical asymptote at $x = a$. To determine which, we evaluate the limit and see that

$$\lim_{x \to a} \frac{3(x-a)(x-b)}{5(x-a)(x-c)} = \lim_{x \to a} \frac{3(x-b)}{5(x-c)} = \frac{3(a-b)}{5(a-c)}\,.$$ This tells us that $R$ has a hole at the point $(a, \frac{3(a-b)}{5(a-c)})$. Moreover, $R$ is not continuous at $x = a$ and $x = c$.

A possible graph of $R$ is shown in the following figure.