Question

Determine the exact mathematical interval(s) in which the function defined by $f(x) = 5x^{3/2} - 3x^{5/2}$ for all $x > 0$ is strictly increasing.

Hint:

Always check the initial domain constraints given in the problem statement. Omitting $x > 0$ might lead to incorrectly including negative values or choosing a closed interval like $(-\infty, 1]$.

Answer 1

Concept: According to the principles of differential calculus, a continuous function $f(x)$ is defined to be increasing on an open interval if its first derivative with respect to $x$, denoted as $f'(x)$, is greater than or equal to zero ($f'(x) \geq 0$) for every point contained within that interval. To find these specific intervals, we find the derivative, locate critical points where $f'(x) = 0$, and perform a sign analysis across the domain.

Step 1: Differentiating the function $f(x)$ with respect to $x$ using the power rule.
The given function is: \[ f(x) = 5x^{3/2} - 3x^{5/2} \quad \text{for } x > 0 \] Let us differentiate this term-by-term using the standard power rule of differentiation, $\frac{d}{dx}(x^n) = n x^{n-1}$: \[ f'(x) = \frac{d}{dx}\left(5x^{3/2}\right) - \frac{d}{dx}\left(3x^{5/2}\right) \] \[ f'(x) = 5 \cdot \left(\frac{3}{2}x^{\frac{3}{2} - 1}\right) - 3 \cdot \left(\frac{5}{2}x^{\frac{5}{2} - 1}\right) \] Simplify the exponents and fractions: \[ f'(x) = \frac{15}{2}x^{1/2} - \frac{15}{2}x^{3/2} \]

Step 2: Factoring the first derivative expression for sign analysis.
To make it easier to analyze the sign of $f'(x)$, let us factor out the greatest common algebraic expression from both terms. The common term is $\frac{15}{2}x^{1/2}$: \[ f'(x) = \frac{15}{2}x^{1/2} \left(1 - x^{3/2 - 1/2}\right) \] \[ f'(x) = \frac{15}{2}\sqrt{x}(1 - x) \]

Step 3: Applying the increasing function condition ($f'(x) \geq 0$).
For the function to be increasing, we must find where: \[ f'(x) \geq 0 \implies \frac{15}{2}\sqrt{x}(1 - x) \geq 0 \] Let us carefully evaluate each factor given the initial domain condition $x > 0$:
• Since $x > 0$, the square root term $\sqrt{x}$ is guaranteed to be a positive real number.
• The constant scalar fraction $\frac{15}{2}$ is also positive. Since the product of positive terms is positive, the overall sign of $f'(x)$ depends entirely on the remaining linear factor $(1 - x)$. Therefore: \[ 1 - x \geq 0 \] Isolating the variable $x$: \[ 1 \geq x \implies x \leq 1 \]

Step 4: Combining the result with the original domain restriction.
The problem states that the function is only defined for $x > 0$. Combining our derived condition $x \leq 1$ with this domain restriction yields: \[ 0 < x \leq 1 \] Expressed in standard interval notation, the function is increasing on the interval $(0, 1]$.