How do you find the critical points for $f (x) = 8 x^{3} + 2 x^{2} - 5 x + 3$ $f(x)=8x^3+2x^2-5x+3$ ?

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Answer 1 · 2015-02-23T23:31:54

Hello,

Calculate the derivative

$f'(x) = 24x^2 + 4x - 5$ .

Solve $f'(x) = 0$ . To do that, Calculate the discriminant $Delta = 4^2 - 4\times 24 times (-5) = 496$ .

The critical points of $f$ are the zeros of $f'$ . So they are
$(-4-sqrt(496))/48$ and $(-4+sqrt(496))/48$ .

You can simplify, because $496 = 16 \times 31$ :
$(-1-sqrt(31))/12$ and $(-1+sqrt(31))/12$ .

How do you find the critical points for f(x)=8x^3+2x^2-5x+3f(x)=8x3+2x2−5x+3f(x)=8x^3+2x^2-5x+3?