Question

max of f(x)= (x-2)^2024 . (3-x)^2025

Answer 1

The maximum value of $f(x)$ is

$$\left(\frac{2024}{4049}\right)^{2024}\left(\frac{2025}{4049}\right)^{2025},$$

attained at

$$x=\frac{10122}{4049}.$$

Answer 2

We wish to maximize

$$f(x)= (x-2)^{2024}(3-x)^{2025}.$$

Since $(x-2)^{2024}$ is always nonnegative and $(3-x)^{2025}$ is positive for $x<3$ (and negative for $x>3$), the maximum must occur for $2<x<3$.

Take the logarithm:

$$\ln f(x) = 2024\ln(x-2) + 2025\ln(3-x).$$

Differentiate with respect to $x$:

$$\frac{d}{dx} \ln f(x) = \frac{2024}{x-2} - \frac{2025}{3-x}=0.$$

Setting the derivative equal to zero gives:

$$\frac{2024}{x-2} = \frac{2025}{3-x}.$$

Cross-multiplying,

$$2024(3-x) = 2025(x-2).$$

Expanding,

$$6072 - 2024x = 2025x - 4050.$$

Collecting like terms,

$$6072 + 4050 = 2025x + 2024x = 4049x,$$

so

$$4049x = 10122 \quad\Longrightarrow\quad x = \frac{10122}{4049}.$$

Now, calculate:

$$x-2 = \frac{10122}{4049} -2 = \frac{10122-8098}{4049} = \frac{2024}{4049},$$ $$3-x = 3 - \frac{10122}{4049} = \frac{12147-10122}{4049} = \frac{2025}{4049}.$$

Thus, the maximum value is

$$f_{\max} = \left(\frac{2024}{4049}\right)^{2024}\left(\frac{2025}{4049}\right)^{2025}.$$