We are given the function $f(x) = \frac{x^2 + A}{x}$, where $A$ is a constant. We also know that $f''(2) = \frac{1}{8}$, and we need to find the value of $A$.

Step 1: Differentiate $f(x)$

Let's start by simplifying the given function: $f(x) = \frac{x^2 + A}{x} = x + \frac{A}{x}$

Now, we will differentiate $f(x)$ to find $f'(x)$ and then $f''(x)$.

First derivative $f'(x)$:

$f'(x) = \frac{d}{dx}\left(x + \frac{A}{x}\right)$ Using the power rule: $f'(x) = 1 - \frac{A}{x^2}$

Second derivative $f''(x)$:

Now, differentiate $f'(x)$: $f''(x) = \frac{d}{dx}\left(1 - \frac{A}{x^2}\right)$ $f''(x) = 0 + \frac{2A}{x^3} = \frac{2A}{x^3}$

Step 2: Use the given condition $f''(2) = \frac{1}{8}$

We are told that $f''(2) = \frac{1}{8}$, so substitute $x = 2$ into the expression for $f''(x)$: $f''(2) = \frac{2A}{2^3} = \frac{2A}{8} = \frac{A}{4}$ Set this equal to the given value $\frac{1}{8}$: $\frac{A}{4} = \frac{1}{8}$

Step 3: Solve for $A$

To solve for $A$, multiply both sides by 4: $A = \frac{4}{8} = \frac{1}{2}$

Thus, the value of $A$ is $\frac{1}{2}$.

Final Answer:

$A = \frac{1}{2}$

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

Here are 5 related questions to explore:

1. How would you differentiate a function involving higher powers of $x$ and constants, like $\frac{x^3 + B}{x}$?
2. Can you solve a similar problem where the given second derivative is $\frac{1}{4}$?
3. How do we apply the chain rule when differentiating more complex functions?
4. What would happen if $A$ were a function of $x$ instead of a constant?
5. How would the solution change if you were asked to find $f'(3)$ instead?

Tip: When solving calculus problems, always simplify functions as much as possible before differentiating—it makes the process quicker and reduces errors.