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 $
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 $
$\lim_{-\infty}f=+\infty$ 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.
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.