Math Problem Statement

For what value of x does the graph of f(x) = e^x - 2x have a horizontal tangent?

Solution

The given problem asks us to find the value of xx at which the graph of f(x)=ex2xf(x) = e^x - 2x has a horizontal tangent. To solve this, we need to find the derivative of the function and set it equal to zero, since a horizontal tangent occurs where the slope (the derivative) is zero.

Step 1: Differentiate the function

The derivative of f(x)=ex2xf(x) = e^x - 2x is:

f(x)=ddx(ex)ddx(2x)=ex2f'(x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(2x) = e^x - 2

Step 2: Set the derivative equal to zero

To find where the tangent is horizontal, set the derivative equal to zero:

ex2=0e^x - 2 = 0

Step 3: Solve for xx

Solve the equation:

ex=2e^x = 2

Take the natural logarithm (ln) of both sides:

x=ln(2)x = \ln(2)

Thus, the value of xx where the graph of f(x)f(x) has a horizontal tangent is x=ln(2)x = \ln(2).

Would you like more details on this problem, or do you have any questions?

Here are five related questions you might find interesting:

  1. What does the function f(x)=ex2xf(x) = e^x - 2x look like graphically?
  2. How does the derivative of an exponential function like exe^x compare to other types of functions?
  3. How can you determine the concavity of the function f(x)f(x) at x=ln(2)x = \ln(2)?
  4. What is the geometric interpretation of a horizontal tangent line?
  5. How can you find the maximum or minimum values of the function f(x)=ex2xf(x) = e^x - 2x?

Tip: When working with exponentials, it's helpful to recall that taking the natural logarithm helps solve equations where the variable is in the exponent.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Exponential Functions

Formulas

f'(x) = e^x - 2

Theorems

The point of horizontal tangency occurs where the derivative equals zero.

Suitable Grade Level

Grades 11-12