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]
