Given () = 2, = 1, and ∆ = 0.6:
a. compute ∆
b. compute
c. illustrate your findings with a sketch in the space provided below

Respuesta :

Answer:

a)

[tex]\triangle y = 1.56[/tex]

b) Dy=1.2

c) See attachment

Step-by-step explanation:

a)Let

[tex] \triangle x[/tex]

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

[tex] \triangle y= f(x + \triangle x) - f(x)[/tex]

We have

[tex]f(x) = {x}^{2} [/tex]

This implies that:

[tex] \triangle y= (x + \triangle x)^{2} - {x}^{2} [/tex]

We substitute

[tex]x = 1 \: and \: \: \triangle x = 0.6[/tex]

This gives

[tex] \triangle y= (1 + 0.6)^{2} - {1}^{2} \\ \triangle y = {1.6}^{2} - 1 = 1.56[/tex]

b) Let

[tex]y = f(x)[/tex]

[tex] \frac{dy}{dx} = f'(x)[/tex]

This implies that:

[tex]dy = f'(x)dx[/tex]

with dx=0.6 , and x=1,

[tex]dy = 2x dx[/tex]

[tex]dy = 2(1)(0.6) = 1.2[/tex]

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