Question

Let f be a differentiable function such that f(3) = 2 and f'(3) = 5. If the tangent line to the graph of f at x = 3 is used to find an approximation to a zero of f, that approximation is?
So confused

Answer 1

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 $x=1.6$.