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Question 1209895: Find the maximum value of f(a), where a is a positive integer, and
f(x) = \frac{13 + x}{2x + 7 - x^2}.
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $f(x) = \frac{13+x}{2x+7-x^2}$. We want to find the maximum value of $f(a)$, where $a$ is a positive integer.
First, let's analyze the denominator: $2x+7-x^2 = -(x^2-2x-7) = -(x^2-2x+1-8) = -(x-1)^2+8 = 8-(x-1)^2$.
The denominator is maximized when $x=1$, and the maximum value is 8.
We need the denominator to be positive, so $8-(x-1)^2 > 0$.
$(x-1)^2 < 8$.
$-\sqrt{8} < x-1 < \sqrt{8}$.
$1-\sqrt{8} < x < 1+\sqrt{8}$.
Since $\sqrt{8} \approx 2.828$, we have $1-2.828 < x < 1+2.828$, so $-1.828 < x < 3.828$.
Since $a$ is a positive integer, $a$ can be 1, 2, or 3.
Now let's evaluate $f(a)$ for $a=1, 2, 3$.
* $f(1) = \frac{13+1}{2(1)+7-1^2} = \frac{14}{2+7-1} = \frac{14}{8} = \frac{7}{4} = 1.75$
* $f(2) = \frac{13+2}{2(2)+7-2^2} = \frac{15}{4+7-4} = \frac{15}{7} \approx 2.143$
* $f(3) = \frac{13+3}{2(3)+7-3^2} = \frac{16}{6+7-9} = \frac{16}{4} = 4$
The maximum value of $f(a)$ is 4, which occurs when $a=3$.
**Therefore, the maximum value of f(a) is 4.**
Answer by greenestamps(13200)  (Show Source):
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