미분 전체 문제

F

2500; the two numbers are each 50.

The minimum sum is $2\sqrt{500}$; the two numbers are each $\sqrt{500}$.

There is no maximum sum; the fundamental equation has only 1 critical value that corresponds to a minimum.

The only critical point of the fundamental equation corresponds to a minimum; to find maximum, we check the endpoints.

If one number is 300, the other number $y$ satisfies $300y = 500$; $y = 5/3$. Thus the sum is $300+5/3$.

The other endpoint, 0, is not feasible as we cannot solve $0\cdot y = 500$ for $y$. In fact, if $0<x<5/3$, then $x\cdot y = 500$ forces $y>300$, which is not a feasible solution.

Hence the maximum sum is $301.\overline{6}$.

Area = 1/4, with sides of length $1/\sqrt{2}$.

The radius should be about $3.84$cm and the height should be $2r=7.67$cm. No, this is not the size of the standard can.

The radius should be about $3.2$in and the height should be $2r=6.4$in. As the #10 is not a perfect cylinder (with extra material to aid in stacking, etc.), the dimensions are close enough to assume that minimizing surface area was a consideration.

The height and width should be $18$ and the length should be $36$, giving a volume of $11,664$in$^3$.

The dog should run about 13 feet along the shore before starting to swim.

The largest area is 2 formed by a square with sides of length $\sqrt{2}$.

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.

A good approximation from the graph is $f(5)=0.5$. We want to find a secant line whose endpoints are both very close to $x=5$, but that also give us clear $y$-values. It looks like $f(5.25) \approx 1$, and $f(4.75)\approx \frac{1}{8}$. The secant line from $x=5$ to $x=5.25$ has approximate slope $\dfrac{f(5.25)-f(5)}{5.25-5}\approx \dfrac{1-.5}{.25}=2$. The secant line from $x=5$ to $x=4.75$ has approximate slope $\dfrac{0.5-\frac{1}{8}}{5-4.75}=\dfrac{3}{2}$.

The graph increases more and more quickly (gets steeper and steeper), so it makes sense that the secant line to the left of $x=5$ has a smaller slope than the secant line to the right of $x=5$. Also, if you're taking secant lines that have endpoints farther out from $x=5$, you'll notice that the slopes of the secant lines change quite dramatically. You have to be very, very close to $x=5$ to get any kind of accuracy.

If we split the difference, we might approximate the slope of the secant line to be the average of $\frac{3}{2}$ and $2$, which is $\frac{7}{4}$.

Another way to try to figure out the tangent line is by carefully drawing it in with a ruler. This is shown here in blue:

It's much easier to take the slope of a line than a curve, and this one looks like it has slope about 1.5. However, we drew this with a computer: by hand it's much harder to draw an accurate tangent line. (That's why we need calculus!)

The actual slope of the tangent line to the function at $x=5$ is about $1.484$. This is extremely hard to figure out just from the graph–by hand, a guess between $1.25$ and $1.75$ would be very accurate.

There is only one tangent line to $f(x)$ at $P$ (shown in blue), but there are infinitely many choices of $Q$ and $R$ (one possibility shown in red). One easy way to sketch the secant line on paper is to draw any line parallel to the tangent line, and choose two intercepts with $y=f(x)$.

Any place the graph looks flat (if you imagine zooming in) is where the tangent line has slope 0. This occurs three times.

Notice that two of the indicated points are at a low point and a high point, respectively. Later, we'll use these places where the tangent line has slope zero to find where a graph achieves its biggest and smallest values.

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.

By definition, $f(x) = x^3$ is differentiable at $x = 0$ if the limit $$\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h^3-0}{h}$$ exists.