Question

Let \( f(x) = 2x^3 - 9x^2 + 7. \) Which of the following is true?

• A. \( f \) is one-one in the interval \([-1, 1]\)
• B. \( f \) is one-one in the interval \([2, 4]\)
• C. \( f \) is NOT one-one in the interval \([-4, 0]\)
• D. \( f \) is NOT one-one in the interval \([0, 4]\)

Hint:

To check one-one nature, use the derivative test: if \( f'(x) \) doesn't change sign in an interval, \( f \) is one-one there.

Answer 1

The Correct Option is D

Step 1: Derivative Calculation

We find the derivative of $f(x)$:

$$f'(x) = \frac{d}{dx} (2x^3 - 9x^2 + 7)$$

$$f'(x) = 6x^2 - 18x$$

Step 2: Critical Points and Sign Analysis

To find the critical points, we set $f'(x) = 0$:

$$6x^2 - 18x = 0$$

$$6x(x - 3) = 0$$

The critical points are $x = 0$ and $x = 3$. These points divide the real number line into three intervals: $(-\infty, 0)$, $(0, 3)$, and $(3, \infty)$.

We analyze the sign of $f'(x)$ in these intervals:

Interval $(-\infty, 0)$: Choose a test value, say $x = -1$.

$$f'(-1) = 6(-1)(-1 - 3) = 6(-1)(-4) = 24 > 0$$

The function $f(x)$ is strictly increasing on $(-\infty, 0]$.

Interval $(0, 3)$: Choose a test value, say $x = 1$.

$$f'(1) = 6(1)(1 - 3) = 6(1)(-2) = -12 < 0$$

The function $f(x)$ is strictly decreasing on $[0, 3]$.

Interval $(3, \infty)$: Choose a test value, say $x = 4$.

$$f'(4) = 6(4)(4 - 3) = 24(1) = 24 > 0$$

The function $f(x)$ is strictly increasing on $[3, \infty)$.

Step 3: Evaluate Options

A function is one-one in an interval if it is strictly monotonic there (i.e., $f'(x)$ does not change sign). A function is NOT one-one if it contains a critical point where the function changes its direction of monotonicity (i.e., $f'(x)$ changes sign) or if it contains a local extremum.

(A) $f$ is one-one in the interval $[-1, 1]$

• The interval $[-1, 1]$ contains the critical point $x=0$.
• In $[-1, 0]$, $f(x)$ is strictly increasing.
• In $[0, 1]$, $f(x)$ is strictly decreasing.
• Since $f(x)$ changes from increasing to decreasing in this interval, it has a local maximum at $x=0$, and thus it is NOT one-one in $[-1, 1]$. (A) is false.

(B) $f$ is one-one in the interval $[2, 4]$

• The interval $[2, 4]$ contains the critical point $x=3$.
• In $[2, 3]$, $f(x)$ is strictly decreasing.
• In $[3, 4]$, $f(x)$ is strictly increasing.
• Since $f(x)$ changes from decreasing to increasing in this interval, it has a local minimum at $x=3$, and thus it is NOT one-one in $[2, 4]$. (B) is false.

(C) $f$ is NOT one-one in the interval $[-4, 0]$

• The function $f(x)$ is strictly increasing on $(-\infty, 0]$, so it is strictly increasing and therefore one-one in the interval $[-4, 0]$. (C) is false.

(D) $f$ is NOT one-one in the interval $[0, 4]$

• The interval $[0, 4]$ contains the critical point $x=3$.
• In $[0, 3]$, $f(x)$ is strictly decreasing.
• In $[3, 4]$, $f(x)$ is strictly increasing.
• Since $f(x)$ changes from decreasing to increasing in this interval, it has a local minimum at $x=3$, and thus it is NOT one-one in $[0, 4]$. (D) is true.

The correct option is (D).