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Hi guys, I was just wondering why bx squared.>4ac

If \displaystyle f(x) = ax - \frac{x^3}{1 + x^2} , show that \displaystyle f'(x) \ge 0 \displaystyle\forall x iff \displaystyle a \ge \frac{9}{8}}

Note that f(x) is a quotient and an inequality is involved. While tempting to use something such as the product rule, by using the quotient rule the denominator will always be > 0.

\displaystyle f'(x) = a - \frac{3x^2(1+x^2) - 2x(x^3)}{(1 + x^2)^2} = a + \frac{x^4 - 3x^2}{(1+x^2)^2} = \frac{(a+1)x^4 + (2a-3)x^2 + a}{(1+x^2)^2}

\displaystyle (a+1)x^4 + (2a-3)x^2 + a \ge 0 \Rightarrow B^2 \le 4AC

\displaystyle (2a - 3)^2 \le 4a(a-1) \Rightarrow a \ge \frac{9}{8}
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YGSK
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Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg
(edited 4 years ago)
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YGSK
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attatchement below
Original post by yash Kainth
Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg


It's not. It's less than or equal to 4ac instead :smile:
Original post by yash Kainth
Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg


Anyway, you want to ensure that for all xx, you have (a+1)x4+(2a−3)x2+a≥0(a+1)x^4 + (2a-3)x^2 +a \geq 0. This means the curve y=(a+1)x4+(2a−3)x2+ay=(a+1)x^4 + (2a-3)x^2 + a must be entirely above the x-axis, or touching it.

If it's entirely above, then there are no roots and we must have the discriminant being less than 0.

If it touches the x-axis, then there are repeated roots and we must have the discriminant being equal to 0.

Combining these case, we require the discriminant to be ≤0\leq 0.
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YGSK
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Thanks, this makes so much more sense now!!

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