To find the equation of the tangent line to the function $f(x) = x \sqrt{x}$ at the point where $x = 4$, follow these steps:

1. Find the function and point of tangency:

We are given: $f(x) = x \sqrt{x} = x \cdot x^{1/2} = x^{3/2}$

At $x = 4$, we can find the corresponding $y$-value: $f(4) = 4 \cdot \sqrt{4} = 4 \cdot 2 = 8$

So, the point of tangency is $(4, 8)$.

2. Find the derivative:

The derivative of the function will give us the slope of the tangent line. Differentiate $f(x) = x^{3/2}$ with respect to $x$:

$f'(x) = \frac{d}{dx} \left( x^{3/2} \right) = \frac{3}{2} x^{1/2}$

3. Evaluate the derivative at $x = 4$:

Now, find the slope of the tangent line by substituting $x = 4$ into the derivative:

$f'(4) = \frac{3}{2} \cdot \sqrt{4} = \frac{3}{2} \cdot 2 = 3$

So, the slope of the tangent line at $x = 4$ is $3$.

4. Write the equation of the tangent line:

The equation of the tangent line is given by the point-slope form: $y - y_1 = m(x - x_1)$ where $m$ is the slope, and $(x_1, y_1)$ is the point of tangency.

Substituting $m = 3$, $x_1 = 4$, and $y_1 = 8$: $y - 8 = 3(x - 4)$

Simplifying this: $y - 8 = 3x - 12$ $y = 3x - 4$

Final Answer:

The equation of the tangent line to $f(x) = x \sqrt{x}$ at $x = 4$ is: $y = 3x - 4$

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Would you like further details on any step or have any questions?

Here are some related questions you might consider:

1. What is the general approach for finding a tangent line to any function?
2. How does the derivative relate to the slope of a tangent line?
3. Can you find the tangent line at $x = 1$ for $f(x) = x \sqrt{x}$?
4. How does the point-slope form of a line work in tangent line problems?
5. What is the geometric interpretation of a tangent line?

Tip: The derivative gives the instantaneous rate of change, or slope, of the function at a specific point.