Question

The interval in which the function \(f(x)=x^2-4x+6\) is increasing, is:

• A. \((2,10)\)
• B. \((2,\infty)\)
• C. \((-2,\infty)\)
• D. \((0,\infty)\)

Hint:

Find \(f'(x)=2x-4\) and solve \(f'(x)>0\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A function is increasing on an interval where its first derivative is positive throughout that interval.
So the task is to find \( f'(x) \) and determine where it is greater than zero.

Step 2: Finding the derivative:
Differentiate \( f(x)=x^2-4x+6 \) with respect to x.
\[ f'(x)=2x-4 \]

Step 3: Solving the inequality:
Set \( f'(x)>0 \) and solve for x.
\[ 2x-4>0 \quad\Rightarrow\quad x>2 \]
Since x can be any real number greater than 2 with no upper bound, the increasing interval is \( (2,\infty) \).

Step 4: Why option A is wrong:
Option A, \( (2,10) \), stops the interval at 10 for no reason; \( f'(x) \) stays positive for all x beyond 10 as well, so the function keeps increasing past 10.

Step 5: Why option C is wrong:
Option C, \( (-2,\infty) \), includes values like \( x=0 \), where \( f'(0)=-4<0 \), so f is actually decreasing there, not increasing.

Step 6: Why option D is wrong:
Option D, \( (0,\infty) \), includes values like \( x=1 \), where \( f'(1)=-2<0 \), so f is decreasing on part of this interval too.

Final Answer:
The derivative is positive only for x greater than 2, so that is the increasing interval. \[ \boxed{(2,\infty)} \]