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Let f(x) = cbrt{x} be the cube root function.
1. Why is this graph said to be symmetric with respect to the origin?
2. Why does this function not have any local minima or local maxima?
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If you put y = x^(1/3) into the MIF function grapher, no negative values show. This is not correct. It shows that graph plotters are not perfect and no substitute for using what you know to consider what the graph looks like.
You cannot find a real number for a negative square root, but you can for all negative values of x.
eg (-8)^(1/3) = -2 .... three negatives multiplied resuts in a negative.
In general if x > 0 and its cube root is a, then the cube root of -x is -a. So for every point (x,a) there is also a point (-x,-a)
This makes it an odd function.
It is an increasing function; ie. its gradient is always positive except at the origin where it is zero.
Using calculus
If you evaluate that power of x by first cube rooting and then squaring the result it will always be positive (except x=0) so this confirms the increasing function.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
If you put y = x^(1/3) into the MIF function grapher, no negative values show. This is not correct. It shows that graph plotters are not perfect and no substitute for using what you know to consider what the graph looks like.
You cannot find a real number for a negative square root, but you can for all negative values of x.
eg (-8)^(1/3) = -2 .... three negatives multiplied resuts in a negative.
In general if x > 0 and its cube root is a, then the cube root of -x is -a. So for every point (x,a) there is also a point (-x,-a)
This makes it an odd function.
It is an increasing function; ie. its gradient is always positive except at the origin where it is zero.
Using calculus
If you evaluate that power of x by first cube rooting and then squaring the result it will always be positive (except x=0) so this confirms the increasing function.
Bob,
You said:
"If you put y = x^(1/3) into the MIF function grapher, no negative values show. This is not correct."
What is not correct?
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Sorry; didn't make myself clear. The grapher doesn't show values of y for negative x; but it should as they certainly exist. So the grapher is not correct.
Not my programming so I cannot do anything about it. As the software is trying to evaluate fractional powers there may be a technical reason why negatives don't get calculated. Does a similar thing happen when you do a fractional power on a calculator?
This is why you need to develop sketching skills.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Sorry; didn't make myself clear. The grapher doesn't show values of y for negative x; but it should as they certainly exist. So the grapher is not correct.
Not my programming so I cannot do anything about it. As the software is trying to evaluate fractional powers there may be a technical reason why negatives don't get calculated. Does a similar thing happen when you do a fractional power on a calculator?
This is why you need to develop sketching skills.
Bob
I never try to input fractional powers on a calculator. I always use Wolframalpha.com for most of my calculator work.
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