Answer
(a) $g'(0)=c+5$ and $g''(0)=c^{2}-2$
(b) $y=(5+3k)x+3$
Work Step by Step
As we are given that
$g(x)=e^{cx}+f(x)$ and $h(x)=e^{kx}f(x)$
where, $f(0)=3, f'(0)=5$ and $f''(0)=-2$
(a) $g'(x)=ce^{cx}+f'(x)$
$g'(0)=ce^{0}+f'(0)=c+5$
This implies
$g'(0)=c+5$
Now, $g''(x)=c^{2}e^{cx}+f''(x)$
$g''(0)=c^{2}e^{0}+f''(0)=c^{2}-2$
Hence, $g'(0)=c+5$ and $g''(0)=c^{2}-2$
(b) Using product rule, we have
$h'(x)=e^{kx}f'(x)+ke^{kx}f(x)$
$h'(0)=e^{0}f'(0)+ke^{0}f(0)=5+3k$
The equation of tangent line to the graph of the function of $h(x)=e^{kx}f(x)$ at x = 0 is given as follows:
$y-h(0)=h'(0)(x-0)$
$y-3=(5+3k).x$
This implies
Hence, $y=(5+3k)x+3$