How do you find the relative extrema for $f (x) = x^{3} - 4 x^{2} + x + 6$ $f(x)=x^3-4x^2+x+6$ ?

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Answer 1 · 2018-07-21T21:39:16

relative min at $(\frac{4}{3} + \frac{\sqrt{13}}{3}, \frac{70}{27} - \frac{26 \sqrt{13}}{27})$ $( 4/3 +(sqrt(13))/3, 70/27 - (26 sqrt(13))/27)$

relative max at $(\frac{4}{3} - \frac{\sqrt{13}}{3}, \frac{70}{27} + \frac{26 \sqrt{13}}{27})$ $(4/3 - (sqrt(13))/3, 70/27 + (26 sqrt(13))/27)$

Given: $f (x) = x^{3} - 4 x^{2} + x + 6$ $f(x) = x^3 - 4x^2 + x + 6$

To find relative extrema first find the first derivative :

$f'(x) = 3x^2 - 8x + 1$

Find the critical value(s) by setting $f'(x) = 0$

$f'(x) = 3x^2 - 8x + 1 = 0$

Use the quadratic formula to find the critical value(s):

$x = (8 +- sqrt(8^2 - 4(3)(1)))/(2*3) = 4/3 +- sqrt(52)/6 = 4/3 +-( sqrt(13))/3$

Find critical points :

$f(4/3 +(sqrt(13))/3) = 70/27 - (26 sqrt(13))/27 ~~-.8794$

$f(4/3 -(sqrt(13))/3) = 70/27 + (26 sqrt(13))/27 ~~6.0646$

Use 2nd derivative test since it is easy to find the 2nd derivative of this function:

$f''(x) = 6x - 8$

If $f''(c) > 0$ we have a relative minimum

If $f''(c) < 0$ we have a relative maximum

$f''(4/3 +(sqrt(13))/3) > 0$ relative min at $x = 4/3 +(sqrt(13))/3$

$f''(4/3 - (sqrt(13))/3) < 0$ relative max at $x = 4/3 - (sqrt(13))/3$

How do you find the relative extrema for f(x)=x^3-4x^2+x+6f(x)=x3−4x2+x+6f(x)=x^3-4x^2+x+6?