I was helping somebody answer some questions when this wild question appears. It looks like this:

Let $f$ be the function given by $f(x)=x^2 - 2x +3$. The tangent line to the graph of $f$ at $x = 2$ is used to approximate values of $f(x)$. What might be the greatest value of $x$, where the error resulting from tangent line approximations is less than $0.5$?

My work

I honestly don't know how to answer it. I found this question when my friend was studying for some licensing exams.....so I turned here for help.

How do you answer the above question?

  • $\begingroup$ As a warmup question, what's the error in the tangent line approximation when $x = 2.1$? $\endgroup$ – littleO Sep 26 '17 at 5:23
  • $\begingroup$ @littleO gotta go find Mr. Google for that oneXD I'll come back when I'm ready..... $\endgroup$ – Palautot Ka Sep 26 '17 at 5:26
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    $\begingroup$ The derivative is defined by $f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$. So, when $x$ is close to $x_0$, we have that $f'(x_0) \approx \frac{f(x) - f(x_0)}{x - x_0}$, or in other words $f(x) \approx f(x_0) + f'(x_0)(x - x_0)$. This function $T(x) = f(x_0) + f'(x_0)(x - x_0)$ provides a good approximation to $f(x)$ when $x$ is near $x_0$. If you graph $T$, its graph is the tangent line to the graph of $f$ at the point $(x_0,f(x_0))$. $\endgroup$ – littleO Sep 26 '17 at 5:32

$(2,3)$ is a touching point and $y-3=2(x-2)$ or $y=2x-1$ is an equation of the tangent.

Thus, $$|x^2-2x+3-(2x-1)|\leq0.5$$ or $$|x-2|\leq\frac{1}{\sqrt2},$$ which gives $$x_{max}=2+\frac{1}{\sqrt2}.$$


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