To find $f(x)$ given the derivative $f'(x) = 1 + 3\sqrt{x}$ and the initial condition $f(4) = 26$, we need to integrate $f'(x)$ and then apply the initial condition to solve for the constant of integration.

Step 1: Integrate $f'(x)$

The derivative $f'(x) = 1 + 3\sqrt{x}$ can be split into two terms for easier integration: $f'(x) = 1 + 3x^{1/2}$ Integrate each term separately: $f(x) = \int \left(1 + 3x^{1/2}\right) \, dx$ $= \int 1 \, dx + \int 3x^{1/2} \, dx$ $= x + 3 \cdot \frac{2}{3} x^{3/2} + C$ $= x + 2x^{3/2} + C$

So, we have: $f(x) = x + 2x^{3/2} + C$

Step 2: Use the Initial Condition $f(4) = 26$

Now, plug in $x = 4$ and $f(4) = 26$ to solve for $C$: $26 = 4 + 2 \cdot 4^{3/2} + C$ Calculate $4^{3/2}$: $4^{3/2} = (2^2)^{3/2} = 2^3 = 8$ So, $26 = 4 + 2 \cdot 8 + C$ $26 = 4 + 16 + C$ $26 = 20 + C$ $C = 6$

Final Solution

Now we can write the function $f(x)$ as: $f(x) = x + 2x^{3/2} + 6$

Answer

$f(x) = x + 2x^{3/2} + 6$

Would you like further details on any part of this solution, or do you have additional questions?

Related Questions

1. How can we verify if $f(x)$ is the correct solution?
2. What if the initial condition was different, such as $f(2) = 10$?
3. How would we integrate if $f'(x)$ included a different power of $x$?
4. How does the constant of integration affect the family of solutions?
5. What techniques are used to solve integrals involving square roots?

Tip

Remember, when integrating functions involving square roots, it's often helpful to rewrite the root as a fractional exponent to simplify the process.