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]
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]