To find the linear approximation of the function $f(x) = \sqrt[5]{x}$ at $a = 1$, we can use the formula for the linear approximation of a function at a point:

$L(x) = f(a) + f'(a)(x - a)$

where:

• $f(a)$ is the value of the function at $x = a$,
• $f'(a)$ is the derivative of $f(x)$ evaluated at $x = a$.

Step 1: Find $f(a)$

For $f(x) = x^{1/5}$, let’s evaluate $f(1)$: $f(1) = 1^{1/5} = 1$

Step 2: Find $f'(x)$

To find $f'(x)$, we differentiate $f(x) = x^{1/5}$ with respect to $x$: $f'(x) = \frac{1}{5} x^{-\frac{4}{5}} = \frac{1}{5x^{4/5}}$

Step 3: Evaluate $f'(a)$ at $x = 1$

$f'(1) = \frac{1}{5 \cdot 1^{4/5}} = \frac{1}{5}$

Step 4: Substitute into the Linear Approximation Formula

Now, we substitute $f(1) = 1$, $f'(1) = \frac{1}{5}$, and $a = 1$ into the formula: $L(x) = f(1) + f'(1)(x - 1)$ $L(x) = 1 + \frac{1}{5}(x - 1)$

Final Answer

The linear approximation of $f(x) = \sqrt[5]{x}$ at $x = 1$ is: $L(x) = 1 + \frac{1}{5}(x - 1)$

Would you like further details on any of these steps or have questions?

Here are some related questions you might find interesting:

1. What is the linear approximation formula and when is it typically used?
2. How do you determine the accuracy of a linear approximation?
3. What is the significance of the point of approximation $a = 1$ in this context?
4. How would the linear approximation change if $f(x)$ had a different exponent?
5. Can we use linear approximation for functions that are not differentiable at a point?

Tip: Linear approximations work best for values of $x$ near the point $a$, as the approximation error increases with distance from $a$.