The problem in the image involves proving that $f(x) \leq f(0)e^x$ for $x > 0$, given the conditions:

1. $f(x)$ is continuous on $[0, +\infty)$,
2. $f(x)$ is differentiable on $(0, +\infty)$,
3. $f'(x) \leq f(x)$ for $x > 0$.

Step-by-step solution:

1. Define an auxiliary function: Let $g(x) = e^{-x}f(x) - f(0)$. This function will help us analyze the behavior of $f(x)$ using its derivative.

2. Evaluate $g(0)$: At $x = 0$, $g(0) = e^{-0}f(0) - f(0) = f(0) - f(0) = 0.$

3. Compute $g'(x)$: Using the product rule and chain rule, the derivative of $g(x)$ is: $g'(x) = \frac{d}{dx}\left(e^{-x}f(x)\right) = (-e^{-x})f(x) + e^{-x}f'(x).$ Therefore: $g'(x) = e^{-x}f'(x) - e^{-x}f(x) = e^{-x}(f'(x) - f(x)).$

4. Analyze $g'(x)$: Given that $f'(x) \leq f(x)$ for $x > 0$, it follows that: $f'(x) - f(x) \leq 0.$ Hence: $g'(x) = e^{-x}(f'(x) - f(x)) \leq 0.$

5. Conclude about $g(x)$: Since $g'(x) \leq 0$, $g(x)$ is a non-increasing function. Furthermore, $g(0) = 0$, so $g(x) \leq 0$ for all $x > 0$.

6. Relate $g(x)$ to $f(x)$: Recall the definition of $g(x)$: $g(x) = e^{-x}f(x) - f(0).$ The inequality $g(x) \leq 0$ implies: $e^{-x}f(x) \leq f(0).$ Multiplying through by $e^x$ (which is positive) gives: $f(x) \leq f(0)e^x \quad \text{for } x > 0.$

Final Answer:

The inequality $f(x) \leq f(0)e^x$ has been proven.

Would you like further clarifications or any additional steps explained?

Related Questions:

1. What happens if $f'(x) = f(x)$? How does this affect the inequality?
2. Can this method be used to analyze other types of differential inequalities?
3. What is the significance of $g(x)$ in this proof? How else could it be defined?
4. Could this proof work for $x < 0$ if $f(x)$ is differentiable there?
5. How does the condition of continuity on $[0, +\infty)$ help in the proof?

Tip:

Always consider defining auxiliary functions when analyzing inequalities involving derivatives. They often simplify and clarify the behavior of the original function.