Linear approximation of $11.2^2$

Question

Use linear approximation, i.e. the tangent line, to approximate $11.2^2$ as follows :

Let $f(x)=x^2$ and find the equation of the tangent line to $f(x)$ at $x = 11$. Using this, find your approximation for $11.2^2$.

So the tangent line is $y = 22x - 121$ and when I insert $x = 11$ in I get $y = 121$, but that is not the correct answer.

Any help?

Answer 1

You should insert $x = 11.2$ in the equation of your tangent line.

Answer 2

dy = f'(x).dx = f'(11) * 0.2 = 22 * 0.2 = 4.4
11.2^2 = f(11) + 4.4 = 121 + 4.4 apx.
11.2^2 = 121 + 2*11*0.2 + 0.04 actual.