Question

The function \[ f(x) = x^{3/5}(5x - 12) \] is increasing in the set:

• A. \( \left( \frac{5}{12}, \infty \right) \)
• B. \( (-\infty, 0) \cup (9, \infty) \)
• C. \( (-\infty, 0) \cup \left( \frac{5}{12}, \infty \right) \)
• D. \( \left( 0, \frac{9}{10} \right) \)
• E. \( \left( \frac{9}{10}, \infty \right) \)

Hint:

To find where a function is increasing or decreasing, compute the first derivative, find the critical points, and check the sign of the derivative on each interval.

Answer 1

Answer: (E)

We are given the function: \[ f(x) = x^{3/5}(5x - 12) \] To find where this function is increasing, we first find its first derivative \( f'(x) \).
Step 1: Differentiate the function
We will use the product rule for differentiation: \[ f'(x) = \frac{d}{dx} \left( x^{3/5} \right) (5x - 12) + x^{3/5} \frac{d}{dx} \left( 5x - 12 \right) \] The derivative of \( x^{3/5} \) is: \[ \frac{d}{dx} \left( x^{3/5} \right) = \frac{3}{5} x^{-2/5} \] The derivative of \( 5x - 12 \) is: \[ \frac{d}{dx} \left( 5x - 12 \right) = 5 \] Thus, the first derivative is: \[ f'(x) = \frac{3}{5} x^{-2/5}(5x - 12) + x^{3/5} \cdot 5 \] Step 2: Set \( f'(x) = 0 \)
To find the critical points, set \( f'(x) = 0 \): \[ \frac{3}{5} x^{-2/5}(5x - 12) + 5x^{3/5} = 0 \] Multiply through by \( 5x^{2/5} \) to eliminate the fractions: \[ 3(5x - 12) + 25x^2 = 0 \] Expanding: \[ 15x - 36 + 25x^2 = 0 \] This simplifies to: \[ 25x^2 + 15x - 36 = 0 \] Step 3: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \( 25x^2 + 15x - 36 = 0 \), we have \( a = 25 \), \( b = 15 \), and \( c = -36 \). Substituting into the quadratic formula: \[ x = \frac{-15 \pm \sqrt{15^2 - 4(25)(-36)}}{2(25)} = \frac{-15 \pm \sqrt{225 + 3600}}{50} = \frac{-15 \pm \sqrt{3825}}{50} \] \[ x = \frac{-15 \pm 61.85}{50} \] Thus, the solutions are: \[ x_1 = \frac{-15 + 61.85}{50} = \frac{46.85}{50} \approx 0.937 \quad {and} \quad x_2 = \frac{-15 - 61.85}{50} = \frac{-76.85}{50} \approx -1.537 \] Step 4: Analyze the intervals
The critical point \( x_1 \approx 0.937 \) (which is approximately \( \frac{9}{10} \)) is where the function changes its behavior. We now test the sign of \( f'(x) \) on the intervals \( \left( \frac{9}{10}, \infty \right) \) and \( (-\infty, \frac{9}{10}) \):
- For \( x>\frac{9}{10} \), \( f'(x)>0 \), so the function is increasing.
- For \( x<\frac{9}{10} \), \( f'(x)<0 \), so the function is decreasing.
Thus, the function is increasing in the interval \( \left( \frac{9}{10}, \infty \right) \).
Thus, the correct answer is option (E), \( \left( \frac{9}{10}, \infty \right) \).