Answer to Question #336019 in Calculus for Valeria

Question

For what values of c does the curve f(x)=2x^3+cx^2+2x have the minimum and maximum points

Answer 1

$f'(x)=6x^2+2cx+2$

"f"(x)=12x+2c"

$f'(x)=0$

$6x^2+2cx+2=0$

$D=4c^2-48$

$x_1=\frac{-2c+\sqrt{4c^2-48}}{12}$

$x_2=\frac{-2c-\sqrt{4c^2-48}}{12}$

$12x(\frac{-2c+\sqrt{4c^2-48}}{12})+2c=0$

$-2c+\sqrt{4c^2-48}=-2c$

$4c^2-48=1$

$c=\sqrt{49/4}=\plusmn 3.5$

If c=3.5

$f'(x)=6x^2+7x+2=0$

$D=49-48=1$

$x=\frac{-7 \plusmn \sqrt1}{12}$

$x_1=-0.5$

$x_2=-2/3$

"f"(-0.5)=12(-0.5)+2(3.5)=1>0"

So it is minimum

"f"(-2\/3)=12(-2\/3)+2(3.5)=-1<0"

So it is maximum

If c=-3.5

$f'(x)=6x^2-7x+2=0$

D=49-48=1

$x=\frac{7 \plusmn \sqrt1}{12}$

$x_1=0.5$

$x_2=2/3$

"f"(-0.5)=12(0.5)+2(3.5)=13>0"

So it is minimum

f"(2/3)=12(2/3)+2(3.5)=15>0

So it is minimum