91Ó°ÊÓ

We're given the function \(y = 1 + 2x + xe^x\). To find the tangent line at a point, we need to find the slope of the curve at that point. And for that, we need to find the derivative of the given function, \(y'(x)\). So, let's find the derivative of \(y = 1 + 2x + xe^x\) with respect to \(x\): First, use the Sum Rule which states that the derivative of a sum of functions is the sum of their derivatives: $$y'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(2x) + \frac{d}{dx}(xe^x)$$ Now, find the derivatives of each term individually: - \(\frac{d}{dx}(1) = 0\) (constant) - \(\frac{d}{dx}(2x) = 2\) (linear term) - For \(\frac{d}{dx}(xe^x)\), use Product Rule which states that \((uv)' = u'v + uv'\): Let \(u = x\) and \(v = e^x\). Then, \(u' = 1\) and \(v' = e^x\). So, \(\frac{d}{dx}(xe^x) = 1 \cdot e^x + x \cdot e^x = e^x(1 + x)\) Combining each term: $$y'(x) = 0 + 2 + e^x(1 + x) = 2 + e^x(1 + x)$$

Now that we have the derivative of the function, we can find the slope of the tangent line at \(a = 0\). Simply substitute \(0\) into the derivative we found: $$y'(0) = 2 + e^0(1 + 0) = 2 + 1(1) = 3$$ The slope of the tangent line at \(a = 0\) is \(3\).

To find the coordinates of the point on the curve at \(a = 0\), we substitute \(x = 0\) into our given function: $$y(0) = 1 + 2(0) + 0e^0 = 1$$ So, the point of tangency is \((0, 1)\).

Now we have the slope of the tangent line (3) and the point of tangency (0, 1). We can use the point-slope form of a line to find the equation of the tangent line: $$y - y_1 = m(x - x_1)$$ Substitute the point \((0, 1)\) and the slope \(3\): $$y - 1 = 3(x - 0)$$ Simplify the equation: $$y = 3x + 1$$ The equation of the tangent line is \(y = 3x + 1\).

Finally, use a graphing software or graphing calculator to plot both the given function, \(y = 1 + 2x + xe^x\), and the tangent line we found, \(y = 3x + 1\), on the same set of axes. You can notice how the tangent line touches the curve at \(a = 0\) and has the same slope as the curve at that point.