8. maximum mark: 4 let l be the tangent to the curve ( y = x e ^ { 3 x } ) at the point ( ( 1, e ^ { 2 } )…

Explanation:

Step1: Find the derivative of the function

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x), (u^\prime=1), (v = e^{3x}), (v^\prime = 3e^{3x}). So (y^\prime=(x)^\prime e^{3x}+x(e^{3x})^\prime=e^{3x}+3xe^{3x}=(1 + 3x)e^{3x}).

Step2: Find the slope of the tangent at (x = 1)

Substitute (x = 1) into (y^\prime). (y^\prime|_{x = 1}=(1+3\times1)e^{3\times1}=4e^{3}).

Step3: Write the equation of the tangent line

Use the point - slope form (y - y_1=m(x - x_1)), where (x_1 = 1), (y_1=e^{3}), (m = 4e^{3}). (y-e^{3}=4e^{3}(x - 1)), which simplifies to (y=4e^{3}x-4e^{3}+e^{3}=4e^{3}x-3e^{3}).

Step4: Find the (x) - intercept

Set (y = 0) in the tangent line equation (0=4e^{3}x-3e^{3}). Solve for (x): [ \begin{align*} 4e^{3}x&=3e^{3}\ x&=\frac{3}{4} \end{align*} ]

Answer:

The coordinates of the point where the tangent meets the (x) - axis is ((\frac{3}{4},0))