SOLUTION: (2x^2-14x+24)/(x^2+6x-40) Max/min values increasing/decreasing @

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Question 1130763: (2x^2-14x+24)/(x^2+6x-40)
Max/min values
increasing/decreasing @

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%282x%5E2-14x%2B24%29%2F%28x%5E2%2B6x-40%29
2%28x%5E2-7x%2B12%29%2F%28x%5E2%2B6x-40%29
2%28%28x+-+3%29+%28x+-+4%29%29%2F%28%28x+%2B+10%29+%28x+-+4%29%29........we will cancel out x-4, but need to remember to exclude x=4 from solution because it makes denominator zero
2%28%28x+-+3%29+cross%28%28x+-+4%29%29%29%2F%28%28x+%2B+10%29cross%28+%28x+-+4%29%29%29
2%28x+-+3%29%2F%28x+%2B+10%29+ -> (for x%3C%3E4)
->exclude +x+=-+10+%7D%7D+from+solution+%0D%0A%0D%0Adomain%3A%0D%0A%0D%0A%7B+%7B%7B%7Bx element R : x%3C%3E-10 and x%3C%3E4 }
-10%3Cx%3C4 or x>4
range:
lim%28x-%3E-infinity%2C%282x%5E2-14x%2B24%29%2F%28x%5E2%2B6x-40%29%29=2
lim%28x-%3E-10%2C-%282x%5E2-14x%2B24%29%2F%28x%5E2%2B6x-40%29%29=-infinity
lim%28x-%3E-10%2C%2B%282x%5E2-14x%2B24%29%2F%28x%5E2%2B6x-40%29%29=infinity
lim%28x-%3E4%2C%282x%5E2-14x%2B24%29%2F%28x%5E2%2B6x-40%29%29=1%2F7

so, range is:
{ y+ element R : y%3C%3E1%2F7 and y%3C%3E2 }

no global maxima and no global minima found

+%282x%5E2+-+14+x+%2B+24%29%2F%28x%5E2+%2B+6+x+-+40%29+ is
increasing | on x%3C-10
increasing | on -10%3Cx%3C4
increasing | on x%3E4

Discontinuities:
x+=+-10+(infinite discontinuity)
+x+=+4 (removable discontinuity)