The problem asks to find the linear approximation of the function $f(x) = \sqrt[4]{x}$ at $x = 16$, without expressing the result in slope-intercept form.

Steps to solve:

1. Linear Approximation Formula: The linear approximation of a function $f(x)$ at $x = a$ is given by the equation: $L(x) = f(a) + f'(a)(x - a)$ where $f'(a)$ is the derivative of $f(x)$ evaluated at $x = a$.

2. Find $f(16)$: Since $f(x) = \sqrt[4]{x}$, $f(16) = \sqrt[4]{16} = 2.$

3. Find $f'(x)$: To find the derivative of $f(x) = \sqrt[4]{x}$, we rewrite it as: $f(x) = x^{1/4}.$ Using the power rule for differentiation: $f'(x) = \frac{1}{4}x^{-3/4}.$

4. Evaluate $f'(16)$: Now, substitute $x = 16$ into $f'(x)$: $f'(16) = \frac{1}{4}(16)^{-3/4} = \frac{1}{4 \cdot 8} = \frac{1}{32}.$

5. Linear Approximation: Now we substitute the values of $f(16) = 2$ and $f'(16) = \frac{1}{32}$ into the linear approximation formula: $L(x) = 2 + \frac{1}{32}(x - 16).$ This is the required linear approximation.

Would you like further details on any step? Also, here are 5 related questions you might want to explore:

1. How does the concept of a tangent line relate to linear approximations?
2. What is the significance of the derivative in the linear approximation?
3. How can the linear approximation be used to estimate values near $x = 16$?
4. What would happen to the approximation if you chose a different point, such as $x = 9$?
5. How does the linear approximation compare to other types of approximations, like quadratic?

Tip: When working with power functions like $x^{1/4}$, always remember that their derivatives can often be expressed using the power rule—simplifying calculations significantly!