Question

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

Answer 1

The graph of
`\(f(x) = e^x - 2x\)` has a horizontal tangent when
`\(x = \ln(2)\)` .

Step-by-step explanation:

To find the value of
`\(x\)` for which the graph of
`\(f(x) = e^x - 2x\)` has a horizontal tangent, we need to find where the derivative of the function is equal to zero. The derivative
`\(f'(x)\)` can be found using the rules of differentiation. Taking the derivative of
`\(e^x\),` we get
`\(e^x\)` , and the derivative of
`\(-2x\) is \(-2\).` Setting
`\(f'(x)\)` equal to zero gives the equation
`\(e^x - 2 = 0\)`. Solving for
`\(x\),` we find
`\(x = \ln(2)\).`

At the critical point
`\(x = \ln(2)\)` , the derivative
`\(f'(x)\)`is equal to zero,
`\(\ln(x)\)`indicating that the slope of the tangent line to the graph is zero, and thus, the graph has a horizontal tangent at this point. The natural logarithm function
`\(\ln(x)\)` is the inverse of the exponential function
`\(e^x\), so \(\ln(2)\)` is the value of
`\(x\)` for which
`\(e^x\)` equals 2.

Understanding the critical points and their significance in calculus allows us to identify key features of a function, such as points where the graph has horizontal tangents. The process of finding critical points involves setting the derivative equal to zero and solving for
`\(x\),` providing valuable information about the behavior of the function.