Question

The tangent line to the graph of a function g at the point x = -7 is y = -(3x+25). Find the tangent line to the graph of y = x g(x) at x = -7. Put your answer into slope-intercept form.

I can't seem to find g(7) since I need to use the product rule.

Answer 1

\[
y = x g(x) \\
\frac{dy}{dx} = xg'(x) + g(x) \\
\left[\frac{dy}{dx}\right]_{x=-7} = (-7)g'(-7) + g(-7)
\]The equation of the tangent to g(x) at x = -7 is: y = -(3x+25)
The slope of the tangent is -3 and therefore \(\large g'(-7) = -3\)

When x = -7, y = -(3*(-7) + 25) = -(-21 + 25) = -4.
Thus, (-7, -4) is a point on the tangent AND g(x). \(\large g(-7) = -4\)
\[
\left[\frac{dy}{dx}\right]_{x=-7} = (-7)g'(-7) + g(-7) = (-7)(-3)+(-4) = 21-4=17\\
y = x g(x) \\
\text{ } \\
\left[y\right]_{x=-7} =(-7)g(-7) = (-7)(-4) = 28 \\
\text{ } \\
\text{Equation of tangent is: } y = mx + b \\
y = 17x + b \\
\text{When } x = -7, y = 28 \\
28 = 17(-7) + b \\
28 = -119 + b \\
b = 28+119 = 147 \\
y = 17x + 147
\]