Respuesta :
Answer:
x ∈ [-0.369, 0.678]
Step-by-step explanation:
To prove - [tex]\sqrt[4]{1 + 2x}[/tex] ≈ 1 + [tex]\frac{1}{2}[/tex] x
As given , f(x) = [tex]\sqrt[4]{1 + 2x}[/tex]
As we know that the linear approximation can be defined as :
L(x) = f(a) + f'(a)(x-a) .........(1)
Now,
We have to show that at a = 0 , equation (1) satisfies
Now,
f(0) = [tex]\sqrt[4]{1 + 0}[/tex] = 1
Also,
f'(x) =
[tex]\frac{d}{dx} (\sqrt[4]{1 + 2x} ) = \frac{1}{4}(1 + 2x)^{\frac{1}{4} - 1}\frac{d}{dx}(1 + 2x) ) \\ = \frac{1}{4}(1 + 2x)^{\frac{-3}{4} } (2)\\ = \frac{1}{2}(1 + 2x)^{\frac{-3}{4} }\\[/tex]
∴ we get
f'(x) = [tex]\frac{1}{2}(1 + 2x)^{-\frac{3}{4} }[/tex]
⇒f'(0) = [tex]\frac{1}{2}[/tex]
Now, equation (1) becomes
L(x) = f(0) + f'(0)(x-0)
= 1 + [tex]\frac{1}{2}[/tex] (x)
⇒ [tex]\sqrt[4]{1 + 2x}[/tex] ≈ 1 + [tex]\frac{1}{2}[/tex] x
Hence verified.
Now,
|[tex]\sqrt[4]{1 + 2x}[/tex] - (1 + [tex]\frac{1}{2}[/tex] x ) | ≤ 0.1
⇒ -0.36893 ≤ x ≤ 0.67766
⇒x ∈ [-0.369, 0.678]
