Question

The function $f(x)=2x^{3}-15x^{2}+36x-24$ is strictly decreasing in the interval is

• A. (2,3)
• B. (1,3)
• C. (2,4)
• D. (1,4)
• E. (2,5)

Hint:

Logic Tip: For any upward-opening quadratic inequality $ax^2 + bx + c<0$ (where $a>0$) with real roots $r_1<r_2$, the solution is always the bounded interval $(r_1, r_2)$. You don't need to test numbers if you remember this geometric rule.

Answer 1

The Correct Option is A

Concept:
A differentiable function is strictly decreasing on any interval where its first derivative is strictly less than zero ($f'(x)<0$).
Step 1: Find the first derivative of the function.
Given $f(x) = 2x^3 - 15x^2 + 36x - 24$. Apply the power rule to differentiate term by term: $$f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(15x^2) + \frac{d}{dx}(36x) - \frac{d}{dx}(24)$$ $$f'(x) = 6x^2 - 30x + 36$$
Step 2: Set up the inequality for a decreasing function.
For the function to be strictly decreasing, we require $f'(x)<0$: $$6x^2 - 30x + 36<0$$
Step 3: Simplify and factor the quadratic expression.
Divide the entire inequality by 6 to simplify the coefficients: $$x^2 - 5x + 6<0$$ Find two numbers that multiply to $6$ and add to $-5$. These are $-2$ and $-3$: $$(x - 2)(x - 3)<0$$
Step 4: Determine the valid interval.
The roots of the equation are $x = 2$ and $x = 3$. These roots divide the number line into intervals: $(-\infty, 2)$, $(2, 3)$, and $(3, \infty)$. Because the leading coefficient of $x^2 - 5x + 6$ is positive (a parabola opening upwards), the expression evaluates to a negative number *between* its roots. Therefore, $(x - 2)(x - 3)<0$ holds true when $x$ is between 2 and 3. The interval is $(2, 3)$.