Linear approximation to 1/0.254 The question says: Use linear approximation to approximate $1/0.254$. I know that $1/0.25 = 4$. Where do I proceed from next. Do I subtract $0.004$ from the answer, or what else could I do?
 A: Let's consider the function $f(x)=\frac{1}{x}$.  We can find the tangent line $L(x)$ of $f$ at $x=\frac{1}{4}$, and then evaluate $L(.254)$ to obtain our linear approximation.
See if you can find $L(x)$ and finish the proof.
A: You want to approximate $f(a+\epsilon)$, and you know $f(a)$.  Here $f(x) = \frac1x, a=0.25,$ and $\epsilon = 0.004$.
Linear approximation assumes that $f$ is something like a straight line near $a$. The slope of the line will be $f'(a)$.  So you calculate the equation of a straight line through the point $(a, f(a))$ with slope $f'(a)$, and then you plug in $a+\epsilon$ into this equation to find where the straight line is at $x=a+\epsilon$. 
Does that help?
A: $$ f(x) \approx f(a) + f'(a) (x-a) $$
with $f(x) = 1/x$.
Since we know $1/0.25 = 4$, let's use that and center our approximation at $a = 0.25$. Now $f'(x) = -1/x^2$. At $a = 1/4$, we have that $f'(a) = -16 $. Therefore:
$$ \frac1x \approx 4 - 16 (x-0.25) $$
Now since $x = 0.254$, we get:
$$ \frac{1}{0.254} \approx 4 - 16 (0.254-0.25) $$
$$ \frac{1}{0.254} \approx 4 - 16 (0.004) $$
$$ \frac{1}{0.254} \approx 4 - 0.064 = 3.936 $$
Note that the exact value is about $3.93700787402$, so our estimate is pretty close.
