Question

Let \(A=[a,\infty)\) denotes the domain, then \(f:[a,\infty)\rightarrow B\) which is defined by \(f(x)=2x^{3}-3x^{2}+6\) will have an inverse for the smallest real value of \(a\) if:

• A. $a=0, \ B=[6,\infty)$
• B. $a=2, \ B=[10,\infty)$
• C. $a=1, \ B=[5,\infty)$
• D. $a=-1, \ B=[5,\infty)$

Hint:

A function can only have an inverse on intervals where its derivative does not change sign. On a graph, the critical points $x=0$ and $x=1$ are the local peaks and valleys where the curve turns around. To keep the function invertible all the way to infinity, we must start our domain at or after the final local valley at $x=1$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A function has an inverse if and only if it is bijective (both one-to-one and onto). For a continuous function on a real interval, being one-to-one is equivalent to being strictly monotonic (either strictly increasing or strictly decreasing).
Step 2: Key Formula or Approach:
1. Find the derivative \( f'(x) \).
2. Determine the intervals where \( f'(x) \ge 0 \) or \( f'(x) \le 0 \).
3. Find the smallest \( a \) such that for all \( x \ge a \), the function is monotonic.
Step 3: Detailed Explanation:
Given \( f(x) = 2x^3 - 3x^2 + 6 \).
Differentiate with respect to \( x \): \[ f'(x) = 6x^2 - 6x = 6x(x - 1) \] The critical points are \( x = 0 \) and \( x = 1 \).
Let's check the sign of the derivative: - For \( x<0 \): \( f'(x)>0 \) (Increasing)
- For \( 0<x<1 \): \( f'(x)<0 \) (Decreasing)
- For \( x>1 \): \( f'(x)>0 \) (Increasing)
We need a domain of the form \( [a, \infty) \) where the function is strictly monotonic. Looking at the sign chart, the function starts increasing and staying increasing from \( x = 1 \) onwards.
Thus, the smallest value of \( a \) such that the function is strictly increasing on \( [a, \infty) \) is \( a = 1 \).
Now find the range \( B \): Since the function is strictly increasing for \( x \ge 1 \), the minimum value occurs at the start of the interval: \[ f(1) = 2(1)^3 - 3(1)^2 + 6 = 2 - 3 + 6 = 5 \] As \( x \to \infty \), \( f(x) \to \infty \).
Therefore, the range \( B = [5, \infty) \).
Step 4: Final Answer:
The smallest \( a \) is 1 and the corresponding range \( B \) is \( [5, \infty) \).