#### Answer

$g(x)=\frac{df(a)}{dx}(x-a)+f(a)$

#### Work Step by Step

Given that
$g(x)=m(x-a)+c$
condition 1: $E(a)=0$
and condition 2: $\lim\limits_{x \to a}\frac{E(x)}{x-a}=0$
now
$E(x)=f(x)-g(x)=f(x)-(m(x-a)+c)=(f(x)-m(x-a)-c)$
on applying condition 1 to the above equation:
$E(a)=f(a)-m(a-a)-c=0$
$c=f(a)$
on applying condition 2 to the above equation:
$\lim\limits_{x \to a}\frac{E(x)}{x-a}=0$
$\lim\limits_{x \to a}\frac{(f(x)-m(x-a)-c)}{x-a}=0$
$\lim\limits_{x \to a}\frac{(f(x)-c)}{x-a}-m=0$
$\lim\limits_{x \to a}\frac{(f(x)-f(a))}{x-a}-m=0$
$\frac{df(a)}{dx}-m=0$
$\frac{df(a)}{dx}=m$
on substituting the value of g(x):
$g(x)=\frac{df(a)}{dx}(x-a)+f(a)$