Exercises
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Exercises
Use the given information to find the equation of the tangent line to the function at the given point.
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Find the equation of the tangent line to the following equations at the given point. Graph the function and the tangent line at that point.
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Each of the following limits represents the derivative of a function, f, at a point, $x=a$. Determine both the function and the point.
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For each function below, find the slope of the tangent line at $x = a$.
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Using the information below, slope of the tangent line and the value of the function at the given point.
If the equation of the tangent line to y at $x=2$ is $y=3x-1$, what are $f(2)$ and $f'(2)$.- If the equation of the tangent line to y at $x=-1$ is $y=-x+4$, what are $f(-1)$ and $f'(-1)$.
Sketch the graph of a function that meets the following requirements.
- $f(1) = 0$, $f'(0) = 0$ and f is decreasing to the left of 0.
- $f(0) = 0$, $f'(1) = f'(-1) = 1$.
- $f(\pm 2) = 0$, $f'(0) = 0$, $\lim_\limits{x\to\infty}f(x) = 1$.
- $f(2) = 3$, $f'(-1) = 0$, $f'(1)=2$, $\lim_\limits{x\to-\infty}f(x) = 0$ and $\lim_\limits{x\to\infty}f(x) = \infty$.
Determine which of the following functions are not differentiable at the origin.
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Explorations
For each of the functions below, find the instantaneous rate of change at the given point. Graph the function and the tangent lines at those points and think about the relationship between the sign of the tangent line's slope and the behavior of the function at each point.
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Applications
- Newton's Law of Gravitation says that the force between two objects whose masses are $m_1$ and $m_2$ and that are r meters apart is given by $F=G\frac{m_1m_2}{r^2}$ where G is Newton's gravitation constant. Assuming the two masses stay constant, how fast is the force changing when the two objects are 1 million meters apart. How fast is it changing when the distance between the objects is 1 meter? Use $G=6.67430 \times 10^{-37}$.
- The demand, Q, for a product in thousands for a given price, p, is given by $Q(p) = -x^2-6x+40$. It may seem unusual for the demand to increase, at least initially, as the price increases but this is a known psychological phenomenon in marketing. A more expensive product is perceived as being higher quality which can increase demand up to a certain point at which the perceived value no longer matches the asked price and the demand will start to decrease. Determine the rate of change of the demand at $x=a$. Based on your result, at what price do you think the demand stops increasing and starts to decrease.
Icons courtesy of icons8.com
Find the equation of the line with the following properties: $y=f(x)$ at $x=2$ if $f(2) = 5$ and $f'(2) = -1$
Remember that the derivative is the slope and $f(2) = 5$ gives you the coordinates of a point on the line.
Find the equation of the line with the following properties: $y=f(x)$ at $x=2$ if $f(2) = 5$ and $f'(2) = -1$
The derivative gives us the slope, $m=f'(2) = -1$ and the equality $f(2) = 5$ tells us that the point $(2, 5)$ is on the line. Substituting those values into the point-slope equation gives us
$$y-5=-1(x-2)$$ $$y - 5 = -x + 2$$ $$y = -x + 7$$Find the equation of the line with the following properties: $y=f(x)$ at $x=-1$ if $f(-1) = 2$ and $f'(-1) = 0$
Remember that a line who's slope is 0 is horizontal.
Find the equation of the line with the following properties: $y=f(x)$ at $x=-1$ if $f(-1) = 2$ and $f'(-1) = 0$
Since the slope of the line is 0, it must be horizontal and must pass through the point $(-1, 2)$. That line is $y=2$.
Find the equation of the tangent line to the equations $f(x)=x^2+2$ at the point $(1, 3)$ then graph the function and the tangent line at that point.
Start by using the definition of the derivative to find the derivative at the point $x = 1$. That will be the tangent line's slope.
Find the equation of the tangent line to the equations $f(x)=x^2+2$ at the point $(1, 3)$ then graph the function and the tangent line at that point.
First we need to find the slope at that point, i.e. the derivative:
$$\begin{aligned} m = f'(1) &= \lim_\limits{h \to 0}\frac{f(h+1)-f(1)}{h} \\ &= \lim_\limits{h \to 0}\frac{(h+1)^2+2-(1^2+2)}{h} \\ &= \lim_\limits{h \to 0}\frac{h^2+2h+1 +2 - 3}{h} \\ &= \lim_\limits{h \to 0}\frac{h^2+2h}{h} \\ &= \lim_\limits{h \to 0}(h+2) \\ &= 2 \end{aligned}$$Now we just need to find the equation of the line with that slope thorugh the point $(1, 3)$:
$$y - 3 = 2(x - 1)$$ $$y = 2x +1$$Find the equation of the tangent line to the equations $f(x)=\frac{1}{x}$ at the point $(1, 1)$ then graph the function and the tangent line at that point.
First we need to find the slope at that point, i.e. the derivative:
$$\begin{aligned} m = f'(1) &= \lim_\limits{x \to 1}\frac{f(x)-f(1)}{x-1} \\ &= \lim_\limits{x \to 1}\frac{\frac{1}{x}-\frac{1}{1}}{x-1} \\ &= \lim_\limits{x \to 1}\frac{\frac{1}{x}-\frac{x}{x}}{x-1} \\ &= \lim_\limits{x \to 1}\frac{\frac{1-x}{x}}{x-1} \\ &= \lim_\limits{x \to 1}\frac{-1}{x} \\ &= -1 \end{aligned}$$Now we just need to find the equation of the line with that slope thorugh the point $(1, 3)$:
$$y - 1 = -1(x - 1)$$ $$y = x + 2$$Find the equation of the tangent line to the equations $f(x)=\sqrt{x+1}$ at the point $(3, 2)$ then graph the function and the tangent line at that point.
$m = \frac{1}{4}$ so the equation is $y=\frac{1}{4}x+\frac{5}{4}$.
The limit $\lim\limits_{h\to 0} \frac{\ln{(h+1)}}{h}$ represents the derivative of a function, f, at a point, $x=a$. Determine both the function and the point.
Think about the definition
$$\lim_\limits{h\to 0}\frac{f(x+a) - f(a)}{h}$$Look for a function in the numerator and something being added to the x will be the value of a.
The limit $\lim\limits_{h\to 0} \frac{\ln{(h+1)}}{h}$ represents the derivative of a function, f, at a point, $x=a$. Determine both the function and the point.
Let $f(x) = \ln(x)$ and $a=1$. Notice that that makes $f(1) = \ln(1) = 0$ which is why there's nothing being subtracted from the logarithm in the numerator. You could think of it as
$$\lim\limits_{h\to 0} \frac{\ln{(h+1) - \ln(1)}}{h} = \lim\limits_{h\to 0} \frac{\ln{(h+1) - 0}}{h} = \lim\limits_{h\to 0} \frac{\ln{(h+1)}}{h}$$The limit $\lim\limits_{h\to 0} \frac{\cos{h}-1}{h}$ represents the derivative of a function, f, at a point, $x=a$. Determine both the function and the point.
Think about the definition
$$\lim_\limits{h\to 0}\frac{f(x+a) - f(a)}{h}$$Look for a function in the numerator and something being added to the x will be the value of a.
The limit $\lim\limits_{h\to 0} \frac{\cos{h}-1}{h}$ represents the derivative of a function, f, at a point, $x=a$. Determine both the function and the point.
Let $f(x) = \cos(x)$ and $a=0$. Notice that that makes $f(h+a) = \cos(h+0) = \cos(h)$ which is why there's nothing being subtracted from the logarithm in the numerator. You could think of it as
$$\lim\limits_{h\to 0} \frac{\cos{(h+0) - \cos(0)}}{h} = \lim\limits_{h\to 0} \frac{\cos h - 1}{h}$$Find the slope of the tangent line to $f(x) = x^2 - 4x + 2$ at $x = a$.
Your final answer should be in terms of a.
Find the slope of the tangent line to $f(x) = \frac{x+1}{x}$ at $x = a$.
$$\begin{aligned} f'(a) &= \lim\limits_{h\to 0} \frac{f(h + a) - f(a)}{h} \\ &= \lim\limits_{h\to 0} \frac{\frac{h+a+1}{h+a} - \frac{a+1}{a}}{h} \\ &= \lim\limits_{h\to 0} \frac{\frac{a(h+a+1)}{a(h+a)} - \frac{(h+a)(a+1)}{a(h+a)}}{h} \\ &= \lim\limits_{h\to 0} \frac{\frac{ah+a^2+a}{a(h+a)} - \frac{a^2+ah+h+a}{a(h+a)}}{h} \\ &= \lim\limits_{h\to 0} \frac{\frac{ah+a^2+a -a^2-ah-h-a}{a(h+a)}}{h} \\ &= \lim\limits_{h\to 0} \frac{\frac{-h}{a(h+a)}}{h} \\ &= \lim\limits_{h\to 0} \frac{-1}{a(h+a)} \\ &= -\frac{1}{a^2} \end{aligned}$$
Find the slope of the tangent line to $f(x) = \frac{2}{x^2}$ at $x = a$.
$$f'(a) = -\frac{4}{a^3}$$
If the equation of the tangent line to y at $x=2$ is $y=3x-1$, what are $f(2)$ and $f'(2)$.
Remember that the slope and the tangent have to meet at $x=2$.
If the equation of the tangent line to y at $x=2$ is $y=3x-1$, what are $f(2)$ and $f'(2)$.
$f'(2)$ is the slope of the tangent line at $x=2$. The slope of the given tangent line is 3 so $f'(2)=3$.
At $x=2$ the y-coordinate of the tangent line is $y=3\cdot2 - 1 = 5$. Since the tangent line and the graph of f touch at that point the function must have the same value so $f(2) = 5$ as well.
Determine if the function $ g(x) = \begin{cases} x^2+1, & x \le 0 \\ 0, & x \gt 0 \end{cases}$ is differentiable at the origin.
You need to determine if $\lim_\limits{h\to 0}\frac{f(h + 0) - f(0)}{h}$ exists but, since $x=0$ is the point where the piecewise defined function splits, you'll have to calculate the left and right hand limits seperately.
Determine if the function $ g(x) = \begin{cases} x^2+1, & x \le 0 \\ 0, & x \gt 0 \end{cases}$ is differentiable at the origin.
We need to determine if $\lim_\limits{h\to 0}\frac{f(h + 0) - f(0)}{h}$ exists but, since $x=0$ is the point where the piecewise defined function splits, we'll have to calculate the left and right hand limits seperately.
| Left Side Limit $$\begin{aligned}\lim_\limits{h\to 0^-}\frac{f(h + 0) - f(0)}{h} &= \lim_\limits{h\to 0^-}\frac{0 - 0}{h} \\ &= \lim_\limits{h\to 0^-} 0 \\ &= 0 \end{aligned}$$ | Right Side Limit $$\begin{aligned}\lim_\limits{h\to 0^+}\frac{f(h + 0) - f(0)}{h} &= \lim_\limits{h\to 0^-}\frac{(h + 0)^2 - 0}{h} \\ &= \lim_\limits{h\to 0^+} \frac{h^2}{h} \\ &= \lim_\limits{h\to 0^+} h \\ &= 0 \end{aligned}$$ |
Since the limit is the same from each side, we can conclude that $f'(0)$ exists and equals 0. This is supported by the graph of the function below but keep in mind that it isn't enough just for the two sides of the function to match up. The requirement is that they have to match up 'smoothly' which you really can't determine from the graph. That's why we do the calculations as well.
Determine if the function $f(x) = \sqrt[3]{x}$ is differentiable at the origin.
$$\begin{aligned} \lim_\limits{h\to 0} \frac{f(h + 0) - f(0)}{h} &= \lim_\limits{h\to 0}\frac{\sqrt[3]{h+0} - \sqrt[3]{0}}{h} \\ &= \lim_\limits{h\to 0}\frac{h^{1/3}}{h} \\ &= \lim_\limits{h\to 0}\frac{1}{h^{2/3}} \\ &= \lim_\limits{h\to 0}\left(\frac{1}{h^{1/3}}\right)^2 \\ &= \infty \end{aligned}$$
I moved the 2 exponent outside the parentheses to emphasize that the expression is always going to be positive. As the denominator goes to 0 the expression goes to plus infinity on the right side and negative infinity on the left but, because of the square, the limit will be plus infinity.
You can see the physical implication of this in the graph below. An infinite or undefined slope means the tangent line is vertical.
Determine if the function $f(x) = \sqrt[3]{x^2}$ is differentiable at the origin.
$$\begin{aligned} \lim_\limits{h\to 0} \frac{f(h + 0) - f(0)}{h} &= \lim_\limits{h\to 0}\frac{\sqrt[3]{(h+0)^2} - \sqrt[3]{0^2}}{h} \\ &= \lim_\limits{h\to 0}\frac{h^{2/3}}{h} \\ &= \lim_\limits{h\to 0}\frac{1}{h^{1/3}} \end{aligned}$$
As h approaches 0 from the left this will be $-\infty$ but as it approaches from the right it will be $+\infty$. Because the two sides are different, the limit doesn't exist so neither does the derivative.
Unlike the $f(x)=\sqrt[3]{x}$ example, you can see from this graph that the function is coming to a 'cusp' at $x=0$ which is a situation where there isn't a unique tangent line so it makes sense that the derivative shouldn't exist.
For the function $f(x) = x^2$ find the instantaneous rate of change at the points $x=-1, 0, 1$. Graph the function and the tangent lines at those points and think about the relationship between the sign of the tangent line's slope and the behavior of the function at each point.
The rates of change are $f'(-1)=2$, $f'(0) = 0$ and $f'(1) = 2$. The point where the derivative is 0 is the minimum value and the slope of the tangent lines one ach side have a different sign.
For the function $f(x) = x^3$ find the instantaneous rate of change at the points $x=-1, 0, 1$. Graph the function and the tangent lines at those points and think about the relationship between the sign of the tangent line's slope and the behavior of the function at each point.
The rates of change are $f'(-1)=3$, $f'(0) = 0$ and $f'(1) = 3$. You've got the same situation here as the previous question where there's a horizontal tangent line but, in this case, the signs of the slopes on each side are the same and we don't have a maximum (or minimum) value.
Newton's Law of Gravitation says that the force between two objects whose masses are $m_1$ and $m_2$ and that are r meters apart is given by $F=G\frac{m_1m_2}{r^2}$ where G is Newton's gravitation constant. Assuming the two masses stay constant, how fast is the force changing when the two objects are 1 million meters apart. How fast is it changing when the distance between the objects is 1 meter? Use $G=6.67430 \times 10^{-37}$.
G and the two masses are all constants so you can factor them out of the limit just like you could with numbers. Don't try replacing G with a number unless you absolutely have to and, if you do, put it off as long as possible. The single letter will be a lot easier to move around than the scientific notation value.
Newton's Law of Gravitation says that the force between two objects whose masses are $m_1$ and $m_2$ and that are r meters apart is given by $F=G\frac{m_1m_2}{r^2}$ where G is Newton's gravitation constant. Assuming the two masses stay constant, how fast is the force changing when the two objects are 1 million meters apart. How fast is it changing when the distance between the objects is 1 meter? Use $G=6.67430 \times 10^{-37}$.
Keep in mind that the only variable in the function is the r - everything else is a constant so we can treat it like we would a number.
$$\begin{aligned} F'(10^6) &= \lim_\limits{h\to 0}\frac{F(h + 10^6)-F(10^6)}{h} \\ &= \lim_\limits{h\to 0}\frac{\frac{Gm_1m_2}{(h + 10^6)^2}-\frac{Gm_1m_2}{(10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{1}{(h + 10^6)^2}-\frac{1}{(10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{(10^6)^2}{(10^6)^2(h + 10^6)^2}-\frac{(h + 10^6)^2}{(h + 10^6)^2(10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{(10^6)^2-(h + 10^6)^2}{(10^6)^2(h + 10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{10^{12}-(10^{12} + 2h\cdot 10^{6} + h^2}{(10^6)^2(h + 10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{10^{12}-10^{12} - 2h\cdot 10^{6} - h^2}{(10^6)^2(h + 10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{-2h\cdot 10^{6} - h^2}{(10^6)^2(h + 10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{\frac{h(-2\cdot 10^{6} - h)}{(10^6)^2(h + 10^6)^2}}{h} \\ &= Gm_1m_2\lim_\limits{h\to 0}\frac{(-2\cdot 10^{6} - h)}{(10^6)^2(h + 10^6)^2} \\ &= Gm_1m_2 \frac{(-2\cdot 10^{6})}{(10^6)^2 \cdot (10^6)^2} \\ &= \frac{-2Gm_1m_2}{(10^6)^3} \\ &= \frac{-2Gm_1m_2}{10^{18}} \end{aligned}$$At this point, you could replace the G with its numeric value but, since we don't have values for the masses, there would be little value in doing that.
To answer the question for $r=1$, you could redo teh calculations from scratch but, if you doing that would mean replacing the $10^6$ with 1. If you just follow through the calculations, replacing each power of 10 with a 1, you'll see that we end up with the same result but with 1 in the pace of the $10^6$ making the final result, $F'(1) = \frac{-2Gm_1m_2}{1^3} = -2Gm_1m_2$.
Remember that that isn't telling us what the force is but how fast the force is changing.
