To find t<span>he relative maximum value of the function we need to find where the function has its first derivative equal to 0.
Its first derivative is -7*(2x)/(x^2+5)^2
</span>7*(2x)/(x^2+5)^2 =0 the numerator needs to be eqaul to 0
2x=0
x=0
g(0) = 7/5
The <span>relative maximum value is at the point (0, 7/5).</span>
Answer:
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Answer:
(x, y) = (5, 1)
Step-by-step explanation:
To <em>eliminate</em> x, you can double the second equation and subtract the first.
... 2(x +4y) -(2x -3y) = 2(9) -(7)
...11y = 11 . . . . . simplify
... y = 1 . . . . . . divide by 11
Using the second equation to find x, we have ...
... x + 4·1 = 9
... x = 5 . . . . . subtract 4
_____
<u>Check</u>
2·5 -3·1 = 10 -3 = 7 . . . . agrees with the first equation
(Since we used the second equation to find x, we know it will check.)
