Question

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

Short answer

The curve is horizontal at \(x = \ln 2\).

Step-by-step solution

Differentiate the given function

Differentiate the given function with respect to\(x\), as shown follows:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {{e^x} - 2x} \right)\\f'\left( x \right) &= \frac{d}{{dx}}\left( {{e^x}} \right) - \frac{d}{{dx}}\left( {2x} \right)\\f'\left( x \right) &= {e^x} - 2\end{aligned}\)

Therefore, \(f'\left( x \right) = {e^x} - 2\).

Apply the condition of Horizontal tangent

The horizontal tangents occur at those values of\(x\)at which\(f'\left( x \right) = 0\). So, set the obtained expression of\(f'\left( x \right)\)equal to\(0\)and solve for\(x\)as follows:

\(\begin{aligned}{e^x} - 2 &= 0\\{e^x} &= 2\\x &= \ln 2\end{aligned}\).

Thus, the curve is horizontal at \(x = \ln 2\).