Question

Find the x-intercept of the tangent line to the graph of f at the point P(0,f(0)) when f(x)= e^-x(5cosx+2sinx)

Answer 1

find
\[f'(x)\] then find
\[f'(0)\] for the slope of the line. the equation will then be
\[y=f'(0)x+f(0)\]

Answer 2

\[f(0)=e^0(5\cos(0)+2\sin(0))=5\]

Answer 3

\[f'(x)=-e^{-x}(5\cos(x)+2\sin(x))+e^{-x}(-5\sin(x)+2\cos(x))\] by the product rule
so
\[f'(0)=-5+2=-3\] if my arithmetic is correct

Answer 4

and therefore the equation for the line is
\[y=-3x+5\]
and so the x- intercept is
\[0=-3x+5\]
\[x=\frac{5}{3}\]