The function f(x) charges value when x changes from x_0 to x_0 + dx. 1(x) = 4x^3

Question

The function f(x) charges value when x changes from x_0 to x_0 + dx. 1(x) = 4x^3 - 4x, x_0 = 1 dx = 01 Find the change Delta f - l(x_0 + dx) - l(x_0) Find the value of the estimate of -f(x_0)dx. Find the approximation error |Delta t - dt| Find the change Delta^2 - f(x_0 - dx) - f(x_0) Delta f = (Round to three decimal places as needed.) Find the value of the estimate dt = f(x_0) dx. df = (Round to three decimal places as needed.) Find the approximation error |Delta f - df| |Delta t - dt| = (Round to three decimal places as needed.)

Explanation / Answer

given f(x)=4x3-4x ,x0=1, dx =0.1

a)f(x0)=4*13-4*1

f(x0)=0

f(x0+dx)=4(1+0.1)3-4(1+0.1)

f(x0+dx)=0.924

f =f(x0+dx)-f(x0)

f =0.924

b) f(x)=4x3-4x

df/dx =12x2 -4

df=(12x2 -4)dx

at xo=1, dx =0.1

df=(12*12 -4)*0.1

df=0.800

c)approximation error =|f-df|

approximation error =|0.924-0.8|

approximation error =0.124

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