Question

find an equation for the line tangent to the graph of $f(x)=-2xe^{x}$ at the point $(a,f(a))$ for $a = 3$.

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the product - rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=-2x\) and \(v = e^{x}\). \(u^\prime=-2\) and \(v^\prime=e^{x}\), so \(f^\prime(x)=-2e^{x}-2xe^{x}=-2e^{x}(1 + x)\).

Step2: Evaluate \(f(3)\)

Substitute \(x = 3\) into \(f(x)=-2xe^{x}\), then \(f(3)=-2\times3\times e^{3}=-6e^{3}\).

Step3: Evaluate \(f^\prime(3)\)

Substitute \(x = 3\) into \(f^\prime(x)=-2e^{x}(1 + x)\), then \(f^\prime(3)=-2e^{3}(1 + 3)=-8e^{3}\).

Step4: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=3\), \(y_{1}=-6e^{3}\) and \(m=-8e^{3}\). So \(y+6e^{3}=-8e^{3}(x - 3)\).

Step5: Simplify the equation

$$
LATEXBLOCK0
$$

Answer:

\(y=-8e^{3}x + 18e^{3}\)