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Find the range of `f(x)=(x^2+34 x-71)/(x^3+2x-7)` |
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Answer» Correct Answer - `(-oo, 5] cup [9,oo)` Let `(x^(2)+34x-71)/(x^(2)+2x-7)=y` or ` x^(2)(1-y)+2(17-y)x+(7y-71)=0` For the real value of x, `b^(2)-4ac ge0` or `4(17-y)^(2)-4(1-y)(7y-71) ge 0` or ` y^(2)-14y+45 ge 0` i.e., `y le 5 " or " y ge 9` Hence, range is `(-oo,5] cup [9,oo).` |
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