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Now i know how to find the max and min local values using an interval if given, but in this question i am not given an interval. How do i go solving the min and max values without it?

$$f(x) = 4 + 6x^2 − 4x^3 $$

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It's local maxima and minima, not "maximal and minima local values". The former refers to values that are, locally, greatest or smallest. The latter would mean local values (which has no meaning in calculus) that are as large or as small as pssible. – Arturo Magidin Jun 27 '12 at 0:25
For a continuous function, local extrema must occur at critical points. The first or second derivative test can be used to determine whether the critical point is a local maximum, a local minimum, or neither. – Arturo Magidin Jun 27 '12 at 0:25
up vote 0 down vote accepted

$\lim_{-\infty}f=+\infty$ and $\lim_{+\infty}f=-\infty$ so your function has no global min or max.

If you want local min/max : $f$ is derivable and $f'(x)=12x-12x^2=12x(1-x)$.
Then $f'(x)<0$ on $(-\infty,0)$, $f'(0)=0$, $f'(x)>0$ on $(0,1)$, $f'(1)=0$ and $f'(x)<0$ on $(1,+\infty)$.
So $f$ is decreasing on $(-\infty,0)$, increasing on $(0,1)$ and decreasing on $(1,+\infty)$.
So $f(0)$ is a local min and $f(1)$ is a local max.

enter image description here

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Where did you get 4 and 6? – soniccool Jun 27 '12 at 2:25
$f(0)=4$ and $f(1)=6$ – JBC Jun 27 '12 at 8:47

Differentiate to get $f'(x) = 12x - 12x^2 = 12x(1-x)$. Draw the sign chart of this function to see that $f'$ is negative for $x < 0$, positive for $x\in(0,1)$, and negative for $x > 1$. Local extrema occur at $x = 0$ and $x = 1$. There are no global extrema since the function increases without bound as $x\to\infty$ and it decreases as $x\to -\infty$ (if you think about moving to the left). You should be able to fill in the details and sketch the graph from this analysis.

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