91Ó°ÊÓ

First, we need to find the derivative of the given function with respect to \(x\). The function is \(y = 1 + 2x + xe^x\). To differentiate the given function, apply the sum rule, product rule, and chain rule: $$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(2x) + \frac{d}{dx}(xe^x)$$ Now we differentiate each term: - \(\frac{d}{dx}(1) = 0\) since the derivative of a constant is 0. - \(\frac{d}{dx}(2x) = 2\) since the derivative of \(x\) is 1. - To differentiate \(xe^x\), we apply the product rule: \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\). Let \(u = x\) and \(v = e^x\). Then, \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = e^x\). So, \(\frac{d}{dx}(xe^x) = x(e^x) + e^x\). Putting it all together, we have: $$\frac{dy}{dx} = 0 + 2 + x(e^x) + e^x$$ which simplifies to: $$\frac{dy}{dx} = 2 + xe^x + e^x$$

Now, we need to find the slope of the tangent line at \(x = 0\). We do this by substituting \(x = 0\) into the derivative we found in step 1: $$\frac{dy}{dx}(0) = 2 + 0(e^0) + e^0 = 2 + 1 = 3$$ The slope of the tangent line at \(x = 0\) is 3.

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

Finally, we use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the curve, and \(m\) is the slope of the tangent line. Using the point \((0, 1)\) and the slope \(m = 3\), the equation of the tangent line is: $$y - 1 = 3(x - 0)$$ Simplifying, we obtain: $$y = 3x + 1$$

Now that we have the equation of the tangent line, the student can use a graphing utility (such as Desmos or Geogebra) to graph both the curve (\(y = 1 + 2x + xe^x\)) and the tangent line (\(y = 3x + 1\)) together on the same set of axes.