If $f(x)=e^{5x}$ and $h(x)=f’’(x)+2f’(x)+f(x)+2$ then $h(0)$ equals
$f(x)=e^{5x}$
$f'(x)=5e^{5x}$
$f''(x)=25 e^{5x}$
$h(x)=25 e^{5x} + 10 e^{5x} + e^{5x} +2$
$h(0)= 25+10+1+2=38$
Option A)
Answer : Option (A)
$ f(x) = e^{5x}$
Differentiating wrt. x,
$f'(x) = 5 \cdot e^{5x}$
$f''(x) = 5 \cdot 5 \cdot e^{5x} = 25 \cdot e^{5x}$
Thus, anwer will be
$25 \cdot e^{0} + 10 \cdot e^0 + e^0 + 2 = 38$