Let's solve this problem step-by-step.

We are given: $f'(x) = \frac{\sqrt{x} - 3}{x^4} \quad \text{for } x > 0$ and the condition $f(1) = 0$. We need to find $f(4)$.

Step 1: Integrate $f'(x)$ to Find $f(x)$

Since $f'(x)$ is the derivative of $f(x)$, we can find $f(x)$ by integrating $f'(x)$ with respect to $x$: $f(x) = \int f'(x) \, dx = \int \frac{\sqrt{x} - 3}{x^4} \, dx$

Rewrite $\sqrt{x}$ as $x^{1/2}$ so we have: $f(x) = \int \frac{x^{1/2} - 3}{x^4} \, dx$

Separate the terms in the integrand: $f(x) = \int \left( \frac{x^{1/2}}{x^4} - \frac{3}{x^4} \right) \, dx = \int \left( x^{-7/2} - 3x^{-4} \right) \, dx$

Step 2: Integrate Each Term

Now we integrate each term separately.

1. For $\int x^{-7/2} \, dx$: $\int x^{-7/2} \, dx = \frac{x^{-7/2 + 1}}{-7/2 + 1} = \frac{x^{-5/2}}{-5/2} = -\frac{2}{5} x^{-5/2}$

2. For $\int -3x^{-4} \, dx$: $\int -3x^{-4} \, dx = -3 \cdot \frac{x^{-4 + 1}}{-4 + 1} = -3 \cdot \frac{x^{-3}}{-3} = x^{-3}$

Thus, $f(x) = -\frac{2}{5} x^{-5/2} + x^{-3} + C$

Step 3: Use the Initial Condition to Find $C$

We know $f(1) = 0$. Substitute $x = 1$ and $f(1) = 0$ into the expression for $f(x)$: $0 = -\frac{2}{5}(1)^{-5/2} + (1)^{-3} + C$ Simplify each term: $0 = -\frac{2}{5} + 1 + C$ $C = -\frac{3}{5}$

So, the function $f(x)$ is: $f(x) = -\frac{2}{5} x^{-5/2} + x^{-3} - \frac{3}{5}$

Step 4: Find $f(4)$

Substitute $x = 4$ into $f(x)$: $f(4) = -\frac{2}{5} \cdot 4^{-5/2} + 4^{-3} - \frac{3}{5}$

Calculate each term:

1. $4^{-5/2} = \frac{1}{32}$, so $-\frac{2}{5} \cdot \frac{1}{32} = -\frac{2}{160} = -\frac{1}{80}$.
2. $4^{-3} = \frac{1}{64}$.

Thus, $f(4) = -\frac{1}{80} + \frac{1}{64} - \frac{3}{5}$

Now, convert each term to have a common denominator (e.g., 320): $f(4) = -\frac{4}{320} + \frac{5}{320} - \frac{192}{320} = \frac{5 - 4 - 192}{320} = \frac{-191}{320}$

So, $f(4) = -\frac{191}{320}$.

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

Related Questions:

1. How do we determine the constant of integration when given an initial value?
2. Why do we rewrite the square root and division terms in terms of exponents?
3. What are the properties of exponents used in this integration?
4. How can we verify our solution by differentiating $f(x)$ again?
5. What would change if $f'(x)$ had a different form?

Tip:

When integrating functions involving powers of $x$, rewrite all terms in terms of exponents to simplify the integration process.