Calculus inflection and concavity? find the points of inflection and discuss the concavity of the graph of the function. f(x)= 2x^3 - 3x^2 -12x 5
Can you take the derivative of this?
yes
its 6x^2-6x-12
ok so do that and set it equal to zero the solutions are your points of inflection.
so 6x^2-6x-12=0
isn't it second derivative?
so solve that
this is what I wanted to confirm because I wasn't sure
no you can also just first take the first derivative to see where the points of inflection are and then take the second derivative to see if its concave up or down at these points.
so can you solve it?
6x^2-6x-12=0 6 (x^2-x-2)
is that right?
now after you are done with that take the second derivative of your equation which is 12x -6
yah i thts right
how do I find the points on that?
i know for sure x=0 right?
no y=0 because its y=6x^2-6x-12
and anyway the solutions for x are -1 and 2
now take the second derivative and plug in these two numbers in it
so 12(-1)-6 and 12(2)-6 so -18 and 18
see how one is positive and the other is negative?
the negative one is concave up and the positive one is concave down
so you can say that sometimes the function is concave up and sometimes its concave down
ok I see. One thing I don't understand though is why did you use y? I thought it was x or f(x)
because f(x) is the y axis
you could also think of it as just f(x)=0
so negative means concave up and positive means concave down
ok I understand. Can I also solve this by finding the second derivative solutions or do i have to find the first equation solutions then insert into second derivative always?>
@mihirb
To me, the first derivative is for critical points, the second derivative for inflection points and concavity only.
@Loser66 this is what I thought
so should I do follow the same steps except just use the second derivative?
i m with you
thank you for clearing that up
XD
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