Use tangent line to find approximation

Question

$f(x) = x^2 - 3x + 5$, the tangent line to the graph of $f$ at $x = 3$ is used to approximate values of $f(x)$.
Which of the following values $3.4$ $3.5$ $3.6$ $3.7$ $3.8$ is the greatest value of x for which the absolute value of the error of approximation from tangent line is less than 0.5?

The answer is $3.7$.
The tangent line I got is $y = 3x - 4$
and at $x = 3.7$, the approximation is $7.1$ while $f(x) = 7.59$
so the result is $\frac {|7.1-7.59|} {7.59} = 0.065$

The problem is that I got similar results for other x-values, they are all far less than $0.5$.
Then what's my mistake?

Answer 1

Okay, let's start with getting the tangent line. The tangent line will be in the form of: $$y=mx+b$$

Where $m$ is the slope, or $f'(3)$. Find the derivative, then:

$$f'(x)=2x-3$$

Hence $f'(3)=3$. Now we have:

$$y=3x+b$$

Plug in $x/y$ to find $b$. Well, we can use $x=3$ and $y=3^2-3(3)+5=5$, so $b$ equals:

$$5=3(3)+b$$ $$b=-4$$

Hence the tangent line's final equation is:

$$y=3x-4$$

Now it's really simple. All you have to is plug in the values of $x$ into the tangent's equation and compare them against the values of the values of $f(x)$. You want to use the Absolute Error for this one, not the percent error! You will do:

$$\Delta x=\big|x_0-x\big|$$

Or, for this case:

$$\text{error: }\big|f(x)-y(x)\big|$$

I made a quick chart to find the answer:

As you can see, the answer is clearly 3.7.