이계도함수와 오목·볼록

Concave up on all $x,$ no inflection points

a. Increasing for all $x$ where defined b. No local minima or maxima c. Concave up for $x<1;$ concave down for $x>1$ d. No inflection points in domain

Since $f'(x)<0$, we need a decreasing function. This only applies to (ii), (iii), and (v). Since $f''(x)>0$, that means $f'(x)$ is increasing, so the slope of the function must be increasing. In (v), the slope is constant, so $f''(x)=0$–therefore, it's not (v). In (iii), the slope is decreasing, because near $a$ the curve is quite flat ($f'(x)$ near zero) but near $b$ the curve is very steeply decreasing ($f'(x)$ is a large negative number), so (iii) has a negative second derivative. By contrast, in (ii), the line starts out as steeply decreasing ($f'(x)$ is a strongly negative number) and becomes flatter and flatter ($f'(x)$ nears 0), so $f'(x)$ is increasing–in other words, $f''(x)>0$. So, (ii) is the only curve that has $f'(x)<0$ and $f''(x)>0$.

In the graph above, the concave-up sections are marked in red. These are where the graph has an increasing derivative; equivalently, where the graph lies above its tangent lines; more descriptively, where it curves like a smiley face.

Concave-down sections are marked in blue. These are where the graph has a decreasing derivative; equivalently, where the graph lies below its tangent lines; more descriptively, where it curves like a frowney face.

The most basic shape of the graph is given by the last two bullet points:

The curve is concave down over the interval $(-5,5)$, so let's give it a frowney-face curvature there.

Finally, when $x>5$ or $x<-5$, our curve should be concave up, so let's give it smiley-face curvature there, without changing its basic increasing/decreasing shape.

This finishes our sketch.

Inflection points occur where $f''(x)$ changes sign. Since $f(x)$ is a polynomial, its first and second derivatives exist everywhere, and are themselves polynomials. In particular,

$$\begin{aligned}f(x)&=3x^5-5x^4+13x\\ f'(x)&=15x^4-20x^3+13\\ f''(x)&=60x^3-60x^2=60x^2(x-1) \end{aligned}$$

The second derivative is negative for $x<1$ and positive for $x>1$. Thus the concavity changes between concave up and concave down at $x=1,\ y=11$.

This is the only inflection point. It is true that $f''(0)=0$, but for values of $x$ both a little larger than and a little smaller than 0, $f''(x)<0$, so the concavity does not change at $x=0$.

(a) Since $f'(2)$ is positive, $f$ is increasing near $x=2$, and since $f''(2)$ is negative, $f$ is concave down near $x=2$. (b) We are given that $f(2) = -3$. Since $f$ is increasing at $x=2$, we expect $f(2.1)$ to be greater than $f(2) = -3$. (c) Since $f''(2)$ is negative, we also know that $f'$ is decreasing near $2$. Thus, we expect that $f'(2.1)$ will be less than $f'(2) = 1.5$. (d) In the following figure, we see a possible graph of a function $f$ that passes through the point $(2,-3)$ at an instantaneous rate of $f'(2) = 1.5$ and in such a way that its second derivative at $x = 2$ is negative.

(a) We will use central differences to estimate the requested derivatives. Thus, $$\begin{aligned}h'(4.5) &\approx \frac{h(5)-h(4)}{5-4} = \frac{69.8 - 55.5}{1} = 14.3 \\ h'(5) &\approx \frac{h(5.5)-h(4.5)}{5.5-4.4} = \frac{81.6 - 60.4}{1} = 21.2 \\ h'(5.5) &\approx \frac{h(6)-h(5)}{6-5} = \frac{93.7 - 69.8}{1} = 23.9\end{aligned}$$ The bungee jumper is rising most rapidly at $t = 5.5$, at a rate of about $23.9$ feet per second. (b) We again use a central difference, this time to estimate $h''(5)$. Observe that $$h''(5) \approx \frac{h'(5.5) - h'(4.5)}{5.5-4.5} = \frac{23.9 - 14.3}{1} = 9.6$$ (c) $h''(5)$ measures the acceleration of the bungee jumper at $t = 5$. Thus we know that the person's acceleration at $t = 5$ is about $9.6$ feet per second per second. This measures the instantaneous rate of change of the jumper's velocity. (d) The bungee jumper's position function is concave down on approximately the intervals $0 < t < 2$, $6 < t < 10$. This is easiest to see if we plot the data using a program like Excel. On these time intervals, the velocity of the jumper is decreasing: either the velocity is positive and getting less positive, or negative and getting more negative.

(a) When a function is increasing, its graph rises as $x$ increases, and when a function is concave up, that means the graph of the function looks bowl-shaped. A graph of an increasing, concave up function $f$ on the interval $(-3,0)$ is shown in the following figure. When a function is concave down, its graph looks like an upside down bowl. The function $f$ in the figure is increasing and concave down on the interval $(0,3)$ Note that $f$ is also increasing on the entire interval $(-3,3)$. (b) When a function is increasing, its graph rises as $x$increases, and when a function is concave down, that means the graph of the function looks like an upside down bowl. A graph of an increasing, concave down function $g$ on the interval $(-3,0)$ is shown in the figure below. When a function is concave up, its graph looks like a bowl. The function $g$ in the figure is increasing and concave up on the interval $(0,3)$. Note that $g$ is also increasing on the entire interval $(-3,3)$. (c) When a function is decreasing, its graph falls as $x$ increases, and when a function is concave up, that means the graph of the function looks bowl-shaped. A graph of a decreasing, concave up function $h$ on the interval $(-3,-1)$is shown in the figure below. To be neither concave up or concave down, a graph has to be linear, so the graph of $h$ in the following figure is a straight line on $(-1,1)$. When a function is concave down, its graph looks like an upside down bowl. The function $h$ in the figure is decreasing and concave down on the interval $(1,3)$. Note that $h$ is decreasing on the entire interval $(-3,3)$. (d) When a function is decreasing, its graph falls as $x$ increases, and when a function is concave down, that means the graph of the function looks like an upside down bowl. A graph of a decreasing, concave down function $p$ on the interval $(-3,0)$ is shown in the figure below. When a function is increasing, its graph rises as $x$ increases. The function $p$ in the figure is increasing and concave down on the interval $(0,3)$.