How do you find the local extrema for $f(x)=(x-3)(x-1)(x+2)$ ?

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Answer 1 · 2017-02-17T22:54:15

We need to find the null points of the first derivative of f(x)

hence

$f'(x) = 3 x^2 - 4 x - 5$

this nullifies at points

$x_1 = 1/3 (2 - sqrt(19))$ and $x_2 = 1/3 (2 +sqrt(19))$

Using the Second-Derivative Test we have that

f(x) has local maximum at $x_1 = 1/3 (2 - sqrt(19))$ with value

$f(1/3 (2 - sqrt(19)))= 2/27 (28 + 19 sqrt(19))$

f(x) has local minimum at $x_1 = 1/3 (2 + sqrt(19))$ with value

$f(1/3 (2 + sqrt(19)))= 2/27 (28 -19 sqrt(19))$

How do you find the local extrema for f(x)=(x-3)(x-1)(x+2)f(x)=(x-3)(x-1)(x+2)?