The maximum and minimum values of x2-14x-9-x2-2x-3 are

The correct option is B 4, 5f(x) = #160; x ^ 2 +14x+9 / x ^ e +2x+3 =1+ 12x+6 / x ^ 2 +2x+3 f'(x)= ( x ^ 2 +2x+3)(12) (12x 16)(2x+2) / ( x ^ 2 +2x+3 ...

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Maxima/Minima of a Quadratic Equation

The maximum a...

Question

The maximum and minimum values of $\frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3}$ are

3, 1

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4, 5

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0, - \infty

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\infty, - \infty

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Solution

The correct option is B 4, 5
f(x) = $\frac{x^{2} + 14 x + 9}{x^{e} + 2 x + 3} = 1 + \frac{12 x + 6}{x^{2} + 2 x + 3}$
$f^{'} (x) = \frac{(x^{2} + 2 x + 3) (12) - (12 x - 16) (2 x + 2)}{(x^{2} + 2 x + 3)^{2}}$
= $\frac{{12 x}^{2} + 24 x + 36 - 24^{2} - 24 x - 12 x - 12}{(x^{2} + 2 x + 3)^{2}}$
= $\frac{12 x^{2} - 12 x + 24}{(x^{2} + 2 x + 3)^{2}}$
= $\frac{- 12 (x^{2} + x - 2)}{(x^{2} + 2 x + 3)^{2}}$
f(x) is decreasing from $(- \infty, - 2) U (1, \infty)$ and increasing from (2, 1)

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The maximum and minimum values of x2+14x+9x2+2x+3 are

The correct option is B 4, 5f(x) = x2+14x+9xe+2x+3=1+12x+6x2+2x+3f′(x)=(x2+2x+3)(12)−(12x−16)(2x+2)(x2+2x+3)2= 12x2+24x+36−242−24x−12x−12(x2+2x+3)2= 12x2−12x+24(x2+2x+3)2= −12(x2+x−2)(x2+2x+3)2f(x) is decreasing from (−∞,−2)U(1,∞) and increasing from (2, 1)

The maximum and minimum values of $\frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3}$ are