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

Answer: (C)

2

Answer 2

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