Find the linear approximation [tex]L(x)[/tex] of the function [tex]g(x)=\sqrt[5]{1+x}[/tex] at [tex]a=0[/tex].

[tex]L(x) \approx \square[/tex]

Use it to approximate the numbers [tex]\sqrt[5]{0.95}[/tex] and [tex]\sqrt[5]{1.1}[/tex]. (Round your answers to three decimal places.)

[tex]
\begin{aligned}
\sqrt[5]{0.95} & \approx \square \\
\sqrt[5]{1.1} & \approx \square
\end{aligned}
[/tex]



Answer :

To find the linear approximation [tex]\( L(x) \)[/tex] of the function [tex]\( g(x) = \sqrt[5]{1 + x} \)[/tex] at [tex]\( a = 0 \)[/tex], we will follow these steps:

1. Determine [tex]\( g(a) \)[/tex]:
Calculate the function value at [tex]\( a = 0 \)[/tex]:
[tex]\[ g(0) = \sqrt[5]{1 + 0} = \sqrt[5]{1} = 1 \][/tex]

2. Compute [tex]\( g'(a) \)[/tex]:
Calculate the derivative of the function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = (1 + x)^{\frac{1}{5}} \][/tex]
Using the power rule for differentiation:
[tex]\[ g'(x) = \frac{1}{5}(1 + x)^{\frac{1}{5} - 1} = \frac{1}{5}(1 + x)^{-\frac{4}{5}} \][/tex]
Evaluate the derivative at [tex]\( x = 0 \)[/tex]:
[tex]\[ g'(0) = \frac{1}{5}(1 + 0)^{-\frac{4}{5}} = \frac{1}{5}(1)^{-\frac{4}{5}} = \frac{1}{5} \][/tex]

3. Linear approximation [tex]\( L(x) \)[/tex]:
The linear approximation of [tex]\( g(x) \)[/tex] at [tex]\( a = 0 \)[/tex] is given by:
[tex]\[ L(x) = g(a) + g'(a) \cdot (x - a) \][/tex]
Plugging in [tex]\( a = 0 \)[/tex]:
[tex]\[ L(x) = g(0) + g'(0) \cdot (x - 0) = 1 + \frac{1}{5} x = 1 + 0.2x \][/tex]
Thus, the linear approximation is:
[tex]\[ L(x) \approx 1 + 0.2x \][/tex]

4. Approximate [tex]\( \sqrt[5]{0.95} \)[/tex] using [tex]\( L(x) \)[/tex]:
[tex]\[ \sqrt[5]{0.95} \approx L(0.95 - 1) \][/tex]
Calculate [tex]\( L(-0.05) \)[/tex]:
[tex]\[ L(-0.05) = 1 + 0.2 \cdot (-0.05) = 1 - 0.01 = 0.990 \][/tex]
Therefore,
[tex]\[ \sqrt[5]{0.95} \approx 0.990 \][/tex]

5. Approximate [tex]\( \sqrt[5]{1.1} \)[/tex] using [tex]\( L(x) \)[/tex]:
[tex]\[ \sqrt[5]{1.1} \approx L(1.1 - 1) \][/tex]
Calculate [tex]\( L(0.1) \)[/tex]:
[tex]\[ L(0.1) = 1 + 0.2 \cdot 0.1 = 1 + 0.02 = 1.020 \][/tex]
Therefore,
[tex]\[ \sqrt[5]{1.1} \approx 1.020 \][/tex]

In conclusion:
[tex]\[ \begin{aligned} L(x) &\approx 1 + 0.2x \\ \sqrt[5]{0.95} &\approx 0.990 \\ \sqrt[5]{1.1} &\approx 1.020 \end{aligned} \][/tex]