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Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no points of intersection. It follows that for some real number \(1

Short Answer

Expert verified
Answer: The approximate value of \(p\) is \(e^{\frac{1}{e}}\), and the intersection point of the functions is approximately \((1.4447, 1.4447)\).

Step by step solution

01

Set up the equations

We are given two functions: \(y=p^x\) and \(y=x\). We need to find the intersection point of these two functions. To do that, we will set them equal to each other and solve the equation, \(p^x = x\). Step 2: Analyze the functions graphically
02

Analyze the functions graphically

By plotting the two functions (\(y=p^x\) and \(y=x\)) on a graph for different values of \(p\), we can visually determine the approximate value of \(p\) that causes the functions to have only one intersection point. Step 3: Determine p value analytically
03

Determine p value analytically

To find the exact value of \(p\), we can take the natural logarithm (ln) of both sides of the equation \(p^x=x\) to get \(\ln(p^x)=\ln(x)\). Applying the power rule for logarithms, we have \(x\ln(p)=\ln(x)\). Now, isolating \(p\) on one side of the equation, we have \(p=e^{\frac{\ln(x)}{x}}\). By analyzing the expression, we find that the value of \(x\) that maximizes the expression for \(p\) is at \(x=e\). Therefore, we can find the value of \(p\) at \(x=e\). Step 4: Calculate the value of p
04

Calculate the value of p

Substitute \(x=e\) into the expression we found in step 3: \(p=e^{\frac{\ln(x)}{x}}\). Calculate \(p=e^{\frac{\ln(e)}{e}}\). Since \(\ln(e)=1\), we have \(p=e^{\frac{1}{e}}\). Step 5: Calculate the intersection point
05

Calculate the intersection point

We now have the value of \(p = e^{\frac{1}{e}}\). To find the intersection point, we need to find the \(x\) and \(y\) values that satisfy both functions. Since both functions are equal at this point, we can plug this value of \(p\) into one of the functions. Let's use the function \(y=p^x\) and plug our value for \(p\): \(y=\left(e^{\frac{1}{e}}\right)^x\) Since the graph equals y=x , we know \(x = y\) at the intersection point. Therefore, substituting y for x: \(y=\left(e^{\frac{1}{e}}\right)^y\) To find the coordinates numerically, we can use numerical methods such as iteration or a calculator to find the approximate value of \(x\) and \(y\) , which are equal. Using a calculator, we find that the intersection point is approximately \((1.4447, 1.4447)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are mathematical expressions of the form \(y = p^x\), where \(p\) is a positive real number known as the base, and \(x\) is the exponent. These functions have a variety of behaviors depending on the value of \(p\):
  • If \(0 < p < 1\), the function is decreasing.
  • If \(p = 1\), it becomes a constant function \(y = 1\).
  • If \(p > 1\), the function is increasing.
Exponential functions are crucial in modeling growth and decay processes, such as population growth, radioactive decay, and interest calculations. They have unique properties that distinguish them from polynomial functions. For instance, exponential functions exhibit rapid growth or decay rates because the rate of change is proportional to the current value, a feature not seen in linear or quadratic functions.
When analyzing exponential functions in conjunction with other functions, such as linear equations like \(y = x\), it can be interesting to explore their interaction points, which quickly lead to the concept of intersection points.
Intersection Points
Intersection points in the context of functions refer to the coordinates where two graphs meet; that is, where they have the same \(y\)-value for the same \(x\)-value. To find the intersection points between \(y = p^x\) and \(y = x\), one must solve the equation \(p^x = x\).
Exploring the case provided, the problem hints that an exponential function \(y = p^x\) intersects the line \(y = x\) at exactly one point for a particular \(p\) between 1 and 2. This scenario suggests that the functions just "touch" each other at some point, creating a tangent between them. The graphical and analytical analysis of this situation often involves plotting the functions to find where this unique point occurs.
By applying analytical techniques, like taking logarithms, we simplify the discovery of the exact \(p\) value. Solving \(p^x = x\) using logarithms yields the condition called the natural logarithm relation: \(x\ln(p) = \ln(x)\). This step helps isolate \(p\) and discover, importantly, that \(p = e^{1/e}\) when \(x = e\), confirming the intersection point.
Graphical Analysis
Graphical analysis is an essential method for understanding the behavior of functions and their intersection points. It involves plotting graphs of functions to visually interpret how they interact with each other. In the case of \(y = p^x\) and \(y = x\), a graphical analysis helps to observe:
  • The variation of intersection points as \(p\) changes.
  • The precise point where the graphs only meet once, indicating tangency.
For \(1 < p < 2\), graphical methods revealed that there's a specific \(p\) where the graph of the exponential function just touches the line \(y = x\), hinting at a tangency intersection. Through plotting, students can observe how the graph \(y = 1.1^x\) intersects twice with \(y = x\), while \(y = 2^x\) does not.
The numerical result obtained through analytical methods (\(p = e^{1/e}\)) can then be confirmed through graphical analysis by plotting the function with this \(p\) and verifying that the intersection occurs at exactly one point, approximately \((1.4447, 1.4447)\). This combined approach provides a more comprehensive understanding and visualization of exponential behaviors and intersections.

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Most popular questions from this chapter

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

Quotient Rule for the second derivative Assuming the first and second derivatives of \(f\) and \(g\) exist at \(x,\) find a formula for \(\frac{d^{2}}{d x^{2}}\left(\frac{f(x)}{g(x)}\right)\)

A woman attached to a bungee cord jumps from a bridge that is \(30 \mathrm{m}\) above a river. Her height in meters above the river \(t\) seconds after the jump is \(y(t)=15\left(1+e^{-t} \cos t\right),\) for \(t \geq 0\). a. Determine her velocity at \(t=1\) and \(t=3\). b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 s. c. Use a graphing utility to estimate the maximum upward velocity.

a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right)\) \(\left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli)

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

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