Why $\Delta y \neq \cos((2.03)^2+1)-2-(\cos(2^2+1)-2)$?

Question

On computing $\Delta y$ from $x=2$ to $x=2.03$:

If $\Delta y = f(x+\Delta x) -f(x)$ and $y=\cos(x^2+1)-x$ why

$\Delta y \neq \cos((2.03)^2+1)-2-(\cos(2^2+1)-2)$ ?

Asumming $\Delta x=0.03$ and $x=2$

Original solution is:

$\Delta y = \cos((2.03)^2+1)-2.03-(\cos(2^2+1)-2)$

This implies:

$\Delta x =0.03$ and $x=2=2.03$?!

Thanks in advance.

$f(x)=\cos(x^2+1)-x \ f(2.03)=\cos((2.03)^2+1)-2.03 \ f(2)=\cos(2^2+1)-2$ And so $\Delta y =f(2.03)-f(2)=[\cos(2.03^2+1)-2.03]-[\cos(2^2+1)-2]$ — randomgirl, Jun 08 '15 at 19:54

Simon S · Accepted Answer · 2015-06-08T19:51:40.657

2

We have $y = f(x) = \cos(x^2 + 1) - x$ and thus

$$\Delta y = f(x+\Delta x) -f(x) = f(2.03) - f(2) = \left(\cos((2.03)^2+1)-\color{red}{2.03}\right)-(\cos(2^2+1)-2)$$

edited Jun 08 '15 at 19:51

answered Jun 08 '15 at 19:47

Simon S

1 Answers1