What are the extrema of $y=2x^3 - 5x^2 - 4x + 7$ ?

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Answer 1 · 2018-04-13T18:08:43

$x=-1/3$ local maximum, $x=5/6$ is turning point and $x=2$ local minimum.

After taking derivative of $y$ and equating to $0$ ,

$6x^2-10x-4=0$

$(3x+1)*(2x-4)=0$ , so $x1=-1/3$ and $x_2=2$

For $x=-1/3$ , $y=140/27$ . So $x=-1/3$ local maximum.

For $x=2$ , $y=-7$ . So $x=2$ local minimum.

After taking second derivative of $y$ and equating to $0$ ,

$12x-10=0$ , so $x=5/6$ is turning point.

What are the extrema of y=2x^3 - 5x^2 - 4x + 7y=2x^3 - 5x^2 - 4x + 7?