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Show 3/2<y<1/2 if y=(x^2+x+1)/(x^2+1)
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You may consider `3/2` as the minimum values of the function and `1/2` the maximum value of the function.
You need to remember that you may evaluate the minimum and the maximum of a function using the derivative of the function, hence, you need to differentiate the function using the quotient rule, such that:
`f'(x) = ((x^2+x+1)'(x^2+1) - (x^2+x+1)(x^2+1)')/((x^2+1)^2)`
`f'(x) = ((2x+1)(x^2+1) - 2x(x^2+x+1))/((x^2+1)^2)`
`f'(x) = (2x^3 + 2x + x^2 + 1 - 2x^3 - 2x^2 - 2x)/((x^2+1)^2)`
Reducing duplicate terms yields:
`f'(x) = (x^2 + 1 - 2x^2)/((x^2+1)^2)`
`f'(x) = (-x^2 + 1)/((x^2+1)^2)`
You need to remember that the function reaches its maximum or minimum at `f'(x) = 0` , hence, you need to solve for x the equation `f'(x) = 0` , such that:
`(-x^2 + 1)/((x^2+1)^2) = 0 => 1 - x^2 = 0 => (1 + x)(1 - x) = 0`
Using the zero product rule yields:
`1 + x = 0 => x = -` 1
`1 - x = 0 => x = 1`
You need to evaluate the function at `x = -1` and `x = 1` , such that:
`f(-1) = (1-1+1)/(1+1) = 1/2`
`f(1) = (1+1+1)/(1+1) = 3/2`
Hence, the maximum value of the function is `f(-1) = 1/2` and the minimum value of the function is `f(1) = 3/2` , thus, the inequality `3/2 < y < 1/2` holds.
Posted by sciencesolve on September 4, 2013 at 4:37 PM (Answer #1)
I believe there may be something wrong with the problem, like a typo or something. For, 3/2 is never less than 1/2.
Posted by steveschoen on September 4, 2013 at 4:53 PM (Answer #2)
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