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Chapter 5: Problem 77

Find a point on the graph of the function \(f(x)=e^{2 x}\) such that the tangent line to the graph at that point passes through the origin. Use a graphing utility to graph \(f\) and the tangent line in the same viewing window.

Short Answer

Expert verified
The point on the graph of the function \(f(x) = e^{2x}\) such that the tangent line at that point passes through the origin is \((2, e^4)\).

Step by step solution

01

Find the derivative of the function

The derivative \(f'(x)\) of the function \(f(x) = e^{2x}\) is \(f'(x) = 2e^{2x}\).
02

Write the equation for the slope

Since the tangent line passes through the origin and the point \((a, f(a))\) on the curve, the slope \(m\) of the tangent is the ratio of the y-coordinate to the x-coordinate. Therefore, \(m = f(a)/a\). Additionally, the slope of the tangent line is equal to the derivative at that point: \(m = f'(a)\).
03

Equate the two expressions for the slope

Next, set the two expressions for the slope equal to each other: \(f(a)/a = f'(a)\). Substituting \(f(a)\) and \(f'(a)\) with their equivalent expressions gives \(e^{2a}/a = 2e^{2a}\).
04

Solve the equation

The equation can be simplified by multiplying both sides by \(a\) and solving for \(e^{2a}\), to get \(a = 2\). The corresponding y-value from the original function is then found by substituting \(a\) into \(f(a)\) to get \(f(2) = e^{2*2} = e^4\).
05

Graph the function and the tangent

Graph the function \(f(x) = e^{2x}\) and its tangent line at the point (2, e^4) in the same viewing window. The tangent line can be drawn using the slope-intercept equation of the tangent line at (a,f(a)), \(y = mx + c\), where \(m = 2e^{2a}\) and \(c=0\) because the line passes through the origin.

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Key Concepts

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

Derivatives
Derivatives are fundamental in calculus and describe how a function changes at a particular point. When you take the derivative of a function, you are calculating the slope of the tangent line to the function at any given point.
This tells us how steep the function is at that point and in what direction it is heading.
For the function \(f(x) = e^{2x}\), the derivative is calculated as follows:
  • The derivative of \(e^{2x}\) is found using the chain rule, which is a method for differentiating compositions of functions.
  • Applying the chain rule gives us \(f'(x) = 2e^{2x}\).
So, \(f'(x) = 2e^{2x}\), which tells us the rate of change of \(f(x)\) at any point \(x\). Understanding derivatives helps us find exact values where the tangent to the curve meets certain conditions, like passing through a specific point such as the origin in this exercise.
Graphing Utilities
Graphing utilities are powerful tools that assist in visualizing the behavior of complex functions. They can plot equations and their derivatives to aid in comprehending how changes affect graph shapes.
In this exercise, after determining the key point at which the tangent passes through the origin, a graphing utility can visualize \(f(x) = e^{2x}\) and its tangent. Here鈥檚 what graphing utilities can be particularly useful for:
  • They provide dynamic interaction with graphs, allowing for close inspection of graph features.
  • Students can see how a tangent line behaves at various points by moving along the curve.
  • Graphing utilities can superimpose multiple functions (like \(f(x)\) and its tangent) for better insight.
This visual aid supports understanding of theoretical solutions by providing concrete, visual evidence of how solutions play out graphically. It reaffirms the calculations of the slope and the point of tangency visually.
Exponential Functions
Exponential functions are a crucial part of mathematics, defined by having a constant base raised to a variable power.
In this problem, the function \(f(x) = e^{2x}\) is exponential, as it involves the constant \(e\), the base of natural logarithms, to the power of a variable expression.
Key properties of exponential functions include:
  • They grow very quickly, which can be seen from their steep slopes.
  • Their graphs are always above the x-axis for all real numbers, as \(e^{x}\) is always positive.
  • They have a constant rate of growth proportional to their current value.
Understanding \(f(x) = e^{2x}\) allows us to see why its graph grows so rapidly and why the slopes are so steep, which are represented by the derivatives we computed. Knowing these properties helps predict and understand the behavior of this type of function in a range of mathematical and real-world scenarios.

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Most popular questions from this chapter

Discuss several ways in which the hyperbolic functions are similar to the trigonometric functions.

Linear and Quadratic Approximations Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a .\) Use a graphing utility to graph the function and its linear and quadratic approximations. $$ f(x)=\cosh x, \quad a=0 $$

Vertical Motion An object is dropped from a height of 400 feet. (a) Find the velocity of the object as a function of time (neglect air resistance on the object). (b) Use the result in part (a) to find the position function. (c) If the air resistance is proportional to the square of the velocity, then \(d v / d t=-32+k v^{2}\), where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Show that the velocity \(v\) as a function of time is \(v(t)=-\sqrt{\frac{32}{k}} \tanh (\sqrt{32 k} t)\) by performing the following integration and simplifying the result. \(\int \frac{d v}{32-k v^{2}}=-\int d t\) (d) Use the result in part (c) to find \(\lim _{t \rightarrow \infty} v(t)\) and give its interpretation. (e) Integrate the velocity function in part (c) and find the position \(s\) of the object as a function of \(t\). Use a graphing utility to graph the position function when \(k=0.01\) and the position function in part (b) in the same viewing window. Estimate the additional time required for the object to reach ground level when air resistance is not neglected. (f) Give a written description of what you believe would happen if \(k\) were increased. Then test your assertion with a particular value of \(k\).

Find any relative extrema of the function. Use a graphing utility to confirm your result. $$ f(x)=\sin x \sinh x-\cos x \cosh x, \quad-4 \leq x \leq 4 $$

Find any relative extrema of the function. Use a graphing utility to confirm your result. $$ g(x)=x \operatorname{sech} x $$
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