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Chapter 3: Problem 42

Let \(f(x)=2 e^{x}-6 x\). a. Find all points on the graph of \(f\) at which the tangent line is horizontal. b. Find all points on the graph of \(f\) at which the tangent line has slope 12.

Short Answer

Expert verified
a. The point with a horizontal tangent on the graph of the function \(f(x)=2e^x-6x\) is \((\ln{3}, 6 - 6\ln{3})\). b. The point with a tangent having a slope of 12 on the graph of the function \(f(x)=2e^x-6x\) is \((\ln{9}, 18 - 6\ln{9})\).

Step by step solution

01

Find the derivative of the function

To find the tangent slope at any point on the curve, we need to differentiate \(f(x)\). We can differentiate \(f(x)=2e^x-6x\) term by term as shown below: \( f'(x) = \frac{d}{dx}(2e^x) - \frac{d}{dx}(6x) = 2e^x - 6\).
02

Find when the tangent is horizontal

To find when the tangent line is horizontal, we set \(f'(x)\) to be 0, and solve for x. \(0 = 2e^x - 6\) Adding 6 to both sides and then dividing by 2, we get: \(e^x = 3\) Now, we take the natural logarithm of both sides: \( \ln{(e^x)} = \ln{3} \Longrightarrow x = \ln{3}\). The x-value at which the tangent is horizontal is \(\ln{3}\), and the corresponding point on the graph is \((\ln{3}, f(\ln{3}))\), which is \((\ln{3}, 2e^{\ln{3}} - 6\ln{3})\) or \((\ln{3}, 6 - 6\ln{3})\).
03

Find when the tangent has a slope of 12

Now, we need to find when the tangent has a slope of 12. To do this, we set \(f'(x)\) equal to 12 and then solve for x. \(12 = 2e^x - 6\) Adding 6 to both sides and then dividing by 2, we get: \(e^x = 9\) Now, we take the natural logarithm of both sides: \( \ln{(e^x)} = \ln{9} \Longrightarrow x = \ln{9}\). The x-value at which the tangent has a slope of 12 is \(\ln{9}\), and the corresponding point on the graph is \((\ln{9}, f(\ln{9}))\), which is \((\ln{9}, 2e^{\ln{9}} - 6\ln{9})\) or \((\ln{9}, 18 - 6\ln{9})\).
04

Answer

a. There is a horizontal tangent at the point \((\ln{3}, 6 - 6\ln{3})\). b. There is a tangent with a slope of 12 at the point \((\ln{9}, 18 - 6\ln{9})\).

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Key Concepts

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

Derivative of Exponential Functions
Understanding the derivative of exponential functions is essential in calculus, especially when dealing with growth and decay problems, or when finding the slope of tangent lines on exponential curves. An exponential function has the form of f(x) = a e^{kx}, where e is the base of the natural logarithm, approximately equal to 2.71828, and a and k are constants.

When you differentiate an exponential function, the rule is fairly straightforward; the derivative of e^{kx} is k e^{kx}. This is powerful because unlike other functions, the derivative of an exponential function is proportional to the function itself. Applying this to our exercise where f(x) = 2e^x - 6x, we find the derivative f'(x) = 2e^x - 6.

This operation is central to finding the rate of change at any point on the curve of f(x), which can represent things like velocity in physics or growth rates in biology. Without knowing how to differentiate these functions, solving many real-world problems in calculus would be significantly harder.
Horizontal Tangent Lines
A horizontal tangent line represents a point on the graph where the slope of the curve is zero. This usually indicates a local maximum or minimum, or a point of inflection, depending on the nature of the function. It's critical for us to find where the derivative of the function equals zero to determine these points.

In the given exercise, we calculate the horizontal tangent by setting the derivative f'(x) = 2e^x - 6 equal to zero and solving for x. This results in the equation e^x = 3, which simplifies to x = ln(3). Consequently, the locus of all points on the graph of the original function f(x) where the tangent line is horizontal can be defined by this specific x-value. This point is pivotal as it represents a stationary point, which is frequently a subject of optimization problems in calculus.
Solving Exponential Equations in Calculus
Solving exponential equations is often necessary when finding where a curve intersects a line or another curve, or identifying points on a graph with specified characteristics, like a particular slope. These types of equations contain variables in the exponent and are solved using properties of exponents and logarithms.

In the case of our problem, solving the exponential equation comes into play when we set the derivative equal to a specific value to find the slope of the tangent line. For the horizontal tangent, we have e^x = 3, which we solve by taking the natural logarithm of both sides, yielding x = ln(3). For a slope of 12, we solve 12 = 2e^x - 6 which simplifies to e^x = 9, and by taking logarithms, we find x = ln(9).

These techniques are part of the foundation students must build to work with more complex functions and to perform more intricate calculus operations, such as finding the area under a curve, determining the volume of a solid of revolution, or solving differential equations that model dynamic systems.

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Most popular questions from this chapter

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