The linear approximation to \(f(x)\) centered at \(x=a\) is
\(f(x)\approx f(a)+f'(a)(x-a)\)
What this means is that you can use the tangent line to \(f(x)\) at \(x=a\) to get a decent approximation of \(f(x)\) at some other value of \(x\) to within a certain degree of accuracy depending on how close this value \(x\) is close to \(a\). (Note that when \(x=a\), the approximation is exact; \(f(a)=f(a)\).)
So what you're asked to do is find an approximate value of a zero of \(f(x)\) near \(a=3\). That is to say, you're looking for some value \(c\) such that \(f(c)=0\), but all you have at your disposal is the linear approximation to the function.
\(f(x)\approx f(3)+f'(3)(x-3)\)
\(f(x)\approx2+5(x-2)\)
\(f(x)\approx5x-8\)
You know that if \(c\) is a zero of \(f\), then \(f(c)=0\), so you get
\(f(c)=0\approx5c-8\)
Solving for \(c\), you find
\(0=5c-8\implies 5c=8\implies c=\dfrac85=1.6\)
This means that an approximate zero of \(f(x)\) is \(x=1.6\).
