To find the absolute maximum value of $f(x) = \frac{25}{x} + 4x$ on the interval $[1, 25]$, we need to:

1. Find the critical points by taking the derivative of $f(x)$ and setting it equal to zero.
2. Evaluate $f(x)$ at the critical points and at the endpoints $x = 1$ and $x = 25$.
3. Compare values to find the maximum.

Step 1: Differentiate $f(x)$

The function is: $f(x) = \frac{25}{x} + 4x$ The derivative $f'(x)$ is: $f'(x) = -\frac{25}{x^2} + 4$

Step 2: Set $f'(x) = 0$ to Find Critical Points

$-\frac{25}{x^2} + 4 = 0$ Solving for $x$: $\frac{25}{x^2} = 4$ $x^2 = \frac{25}{4}$ $x = \pm \frac{5}{2}$ Since $x = \frac{5}{2} = 2.5$ lies within the interval $[1, 25]$, it is a valid critical point.

Step 3: Evaluate $f(x)$ at $x = 1$, $x = 25$, and $x = 2.5$

1. At $x = 1$: $f(1) = \frac{25}{1} + 4 \cdot 1 = 25 + 4 = 29$
2. At $x = 25$: $f(25) = \frac{25}{25} + 4 \cdot 25 = 1 + 100 = 101$
3. At $x = 2.5$: $f(2.5) = \frac{25}{2.5} + 4 \cdot 2.5 = 10 + 10 = 20$

Step 4: Determine the Maximum Value

The values are:

• $f(1) = 29$
• $f(25) = 101$
• $f(2.5) = 20$

Thus, the absolute maximum value on the interval $[1, 25]$ is 101 at $x = 25$.

Would you like further details on any step?

Related Questions:

1. How do we find the absolute minimum value of $f(x)$ on the interval $[1, 25]$?
2. What happens if we change the interval? How does the maximum value shift?
3. How would the procedure change if $f(x)$ were given in a different form?
4. Can we use a graphical approach to verify the maximum?
5. What are other methods to find critical points of a function?

Tip:

Always evaluate both critical points and endpoints when finding absolute extrema on a closed interval.