Prove that y=x^3 is monotonic increasing for all x /= 0.
since y' = 3x^2 > 0 for x≠0,
the slope is always positive. That is, y is always increasing.
To prove that the function y = x^3 is monotonic increasing for all x ≠ 0, we need to show that the derivative of the function is always positive for x ≠ 0.
1. Start with the given function: y = x^3.
2. Take the derivative of the function with respect to x:
dy/dx = 3x^2.
3. Now we need to analyze the sign of the derivative for different values of x:
- For x > 0: The derivative dy/dx = 3x^2 is positive, since x^2 is always positive for x ≠ 0.
- For x < 0: The derivative dy/dx = 3x^2 is also positive, since the square of a negative number is positive.
4. Thus, in both cases, the derivative is positive for all x ≠ 0.
5. Since the derivative is always positive, this implies that the function y = x^3 is monotonic increasing for all x ≠ 0.
Therefore, we have successfully proven that the function y = x^3 is monotonic increasing for all x ≠ 0.
To prove that the function y = x^3 is monotonic increasing for all x ≠ 0, we need to show that for any two values of x, say x1 and x2, such that x1 < x2, we have y(x1) < y(x2).
First, let's evaluate y(x1) = (x1)^3 and y(x2) = (x2)^3:
y(x1) = (x1)^3
y(x2) = (x2)^3
Now, let's substitute the values of x1 and x2 in terms of their inequality, x1 < x2:
y(x1) = (x1)^3
y(x2) = (x2)^3
Since x1 < x2, we can raise the inequality to the power of 3 without changing the direction of the inequality:
(x1)^3 < (x2)^3
So, we have shown that y(x1) < y(x2) for any x1 < x2. Therefore, the function y = x^3 is monotonic increasing for all x ≠ 0.
To generalize this proof, we can also use calculus to show that the derivative of the function y = x^3 is positive for all x ≠ 0. The derivative of y = x^3 with respect to x is given by:
dy/dx = 3x^2
For any x ≠ 0, 3x^2 will always be positive. Therefore, the derivative is positive, indicating that the function y = x^3 is increasing for all x ≠ 0.