Question

For what real values of $a$ and $b$ are all the extrema of the function $f\left(x\right)=\frac{5a^2}{3}(x^3+2a{x}^2-9x+b)$ positive and the maximum is at the point $x_0=-\frac{5}{9}$?

Answer 1

$f^{\prime }\left(x\right)=\frac{5a^2}{3}(3x^2+4ax-9)f^{\prime }\left(\frac{-5}{9}\right)=03\times \frac{25}{81}-4a\times \frac{5}{9}-9=0\frac{-218}{27}=\frac{20a}{9}a=-\frac{109}{30}$

for $f\left(x\right)>0at{x}_0=\frac{-5}{9}$

$(x^3+2a{x}^2-9x+b)>0$

$b>9x-x^3-2a{x}^2$

$=9\times -(\frac{-5}{9}{)}^3+2\times \frac{109}{30}\times (\frac{5}{9}{)}^2$

$b>-2.59$