Question

Given () = 2, = 1, and ∆ = 0.6:

Answer 1

Answer:

a)

$\triangle y = 1.56$

b) Dy=1.2

c) See attachment

Step-by-step explanation:

a)Let

$\triangle x$

be a small change in x, then the corresponding change in y is

$\triangle y= f(x + \triangle x) - f(x)$

We have

$f(x) = {x}^{2}$

This implies that:

$\triangle y= (x + \triangle x)^{2} - {x}^{2}$

We substitute

$x = 1 \: and \: \: \triangle x = 0.6$

This gives

$\triangle y= (1 + 0.6)^{2} - {1}^{2} \\ \triangle y = {1.6}^{2} - 1 = 1.56$

b) Let

$y = f(x)$

$\frac{dy}{dx} = f'(x)$

This implies that:

$dy = f'(x)dx$

with dx=0.6 , and x=1,

$dy = 2x dx$

$dy = 2(1)(0.6) = 1.2$