Tangents at x=0 and intersections with a line

Problem

For each function find the tangent line to the graph at $x=0$ and one of its intersections with the line $y=-\dfrac{x}{3}-3$: a) $f(x)=-5+2e^{-x}$; b) $f(x)=-2\cos x-5$; c) $f(x)=-5\sin(2x)-2$.

Solution

a) $f(0)=-3$, $f'(x)=-2e^{-x}$, $f'(0)=-2$: tangent $y=-2x-3$. Since the line also passes through $(0,-3)$, the intersection is $(0,-3)$. b) $f(0)=-7$, $f'(x)=2\sin x$, $f'(0)=0$: tangent $y=-7$. The intersection with the line ($-2\cos x-5=-\tfrac{x}{3}-3$) is transcendental: $x\approx7{,}57$. c) $f(0)=-2$, $f'(x)=-10\cos(2x)$, $f'(0)=-10$: tangent $y=-10x-2$. The intersection ($5\sin 2x=\tfrac{x}{3}+1$) is $x\approx0{,}10$. $\boxed{\,a)\ y=-2x-3\quad b)\ y=-7\quad c)\ y=-10x-2\,}$