If x is real, the maximum value of 3x^2+9x+173x^2+9x+7 is

Question

If $x$ is real, the maximum value of $\dfrac{3x^2+9x+17}{3x^2+9x+7}$ is

Accepted Answer

Let $y=\dfrac{3x^2+9x+17}{3x^2+9x+7}$: $3x^2(y-1)+9x(y-1)+7y-17=0$. Since $x$ is real, $D\ge0$:

$$81(y-1)^2-12(y-1)(7y-17)\ge0 \implies (y-1)(y-41)\le0 \implies 1\le y\le41.$$

Maximum: $y=41.$

Solutions · 1