x2/3+y 2/3= 1

Step by step solution Step 1 of 2 Given relation x2/+y 2/3= 1. Let’s rearrange this relation: x2/+y 2/3= 1 y 2/3= 1x 2/3 y = (1x )2/33/2 2/3 y = (1x ) Therefore we have the function : 3 y = f(x) = (1x )2/3 This function is square root function and for any square root function the quantity under the root must be nonnegative. This means that: (1x ) 0 1x 2/3 0 1 x 1 (the function (x) is only defined for all real numbers between -1 and 1) Let s find the value (-x): 3 f(x) = (1(x) )/3 2/33 f(x) = (1(x) ) 2/3 3 2 3 2 3 2 If we write (x) as (x) then we can noticed that (x) = (+x) . Therefore we have: f(x) = (1(x) ) =3 (1(+x) )/33 f(x) = f(x) Based on the previous we an conclude that the funct (x) is even and it is symmetric with respect to th -axis. If the function is symmetric with respect to the y-axis then its values for and x are equal. Let’s look the the graph of f(x)..