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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}

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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Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg

Last edited by yash Kainth; 5 months ago

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(Original post by

Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg

**yash Kainth**)Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg

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**yash Kainth**)

Why is b squared greater than 4ac 93DA80FD-2CDA-42F2-8323-AD747B41264C.jpeg

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 .

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