16 / (3(x-4)^(1/3))
How would I go about solving for x in this equation? The (x-4)^(1/3) part is basically the same as the cube root of (x-4) right?
P.S. Sorry I'm totally new at this site and haven't taken the time to learn how to use the commands so my equations are a bit of a mess
Okay, well here's the original equation, but it belongs in the calculus forum. I'm supposed to be finding a single critical point for it:
f(x) = 8(x-4)^(2/3)
and f'(x) = (2/3) 8((x-4)^(-1/3)) (1) ... hopefully I did that derivative right. Maybe that's where I'm screwing up in the first place.
and that equation can be set up as 16 / (3(x-4)^(1/3)... I'm not sure which would be easier to solve.
And then in all the other equations we've done you would solve for x by setting the left side to 0... so basically I have no clue what to do.
No, you are not supposed to find an expression and set it equal to zero. You are supposed to find critical points of the function and gain some understanding.
The Function - You should notice that if this were an integer exponent it would cause you little problem. The complication is the fractional exponent. Okay, why is that different? How does it act differently? Is there a Domain difference? If it were 3/2, would that be different from 2/3? Have you tried to draw it? There is a subtle problem if you want some CAS to draw it for you. Do you know this? You should find that is NOT the same as . Why would that be?
The First Derivative - Why don't we see what we can tell about the function itself before we get into this?