Question

Find an equation of the tangent line to the curve at the given point.

38. \(y = 2{e^x} + x,\;\;\;\;\;\;\;\left( {0,2} \right)\)

Short answer

The required equation is \(y = 3x + 2\).

Step-by-step solution

Find the slope of the equation

Determine the derivative \(\frac{{dy}}{{dx}}\) to represent the slope.

\(\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {2{e^x} + x} \right)\\ &= \frac{d}{{dx}}\left( {2{e^x}} \right) + \frac{d}{{dx}}\left( x \right)\\y' &= 2{e^x} + 1\end{align}\)

Find the slope of the equation at given point

Substitute the given point into the obtained equation of slope to find the slope at particular point.

\(\begin{align}y'\left( 0 \right) &= 2{e^0} + 1\\ &= 2 + 1\\ &= 3\end{align}\)

Determine the tangent line equation

The equation of tangent line is given by \(y - {y_1} = m\left( {x - {x_1}} \right)\), where \(\left( {{x_1},{y_1}} \right)\) is the given pint and \(m\) is the slope at given point.

Substitute the values into the equation to find the equation of tangent line at the point \(\left( {0,2} \right)\).

\(\begin{align}y - 2 &= 3\left( {x - 0} \right)\\y - 2 &= 3x\\y &= 3x + 2\end{align}\)

Thus, the required equation is \(y = 3x + 2\).