Question

The minimum possible value of \(\frac{x^2−6x+10}{3−x}\),for x<3,is

• A. \(\frac{1}{2}\)
• B. \(-\frac{1}{2}\)
• C. 2
• D. -2

Answer 1

The Correct Option is C

The correct answer is C:2
We want to find the minimum value of the expression \(\frac{(x^2 - 6x + 10)}{(3 - x)}\) for x<3. 
Find Critical Points: 
Take the derivative of the expression with respect to x: 
\((\frac{d}{dx}) (\frac{(x^2 - 6x + 10)}{(3 - x)}) \)
Using the quotient rule,simplify the derivative and set it equal to zero: 
\((3 - x)(2x - 6) - (x^2 - 6x + 10)(-1) = 0 \)
This simplifies to the quadratic equation: 
\(3x^2 - 14x + 16 = 0 \)
Solve the quadratic equation using the quadratic formula to find two potential critical points: x=2 and \(x = \frac{4}{3}\). 
Check Interval: 
We need to determine which critical point lies within the interval x<3.Clearly, x=2 satisfies this condition. 
Calculate Minimum Value: 
Plug x=2 back into the expression \(\frac{(x^2 - 6x + 10)}{(3 - x)}\): 
\(\frac{(2^2 - 6\times{2} + 10)}{(3 - 2)} = \frac{(4 - 12 + 10)}{1} = 2 \)
So, the minimum value of the expression for x<3 is 2, and the correct answer is: c. 2

Answer 2

Given \(x<3\) and \(3−x>0\), 
let\( 3−x=y\). Consequently, \(y>0\)
\(\frac{x^2 - 6x + 10}{3 - x}\) now equals \(\frac{x^2 - 6x + 9 + 1}{3 - x}\)

= \(\frac{{(x - 3)^2 + 1}}{{3 - x}}\)
The equation will become \(y^2+\frac{1}{y}\) or \(y+\frac{1}{y}\) as \(3−x=y\).

When y = 1, the expression \(y+\frac{1}{y}\) for \(y > 0\), will have a minimum value of \(1 + 1 = 2 \)
Therefore, choice B is the right one.