What are the intervals where the graph of y equals x3-3x are increasing?

Answer 1

6

Answer 2

I am interpreting this as:

"On what intervals are the function y=x3-3x increasing?"

This function can also be written as f(x)=x3-3x

To find out, you must examine the first derivative of the function. The first derivative gives the slope of the lines tangent to the function at any value of x. By using the power rule, you can get the derivative to be:

y'=(3)(1)x3-1-(3)(1)x1-1=3x2-3(1)=3x2-3.

This can also be written as f'(x)=3x2-3

To find where the function is increasing, you must first find where the derivative is zero. If a function is decreasing, its first derivative is negative. If it is increasing, the first derivative is positive. So, it follows that for a function to change from increasing to decreasing or from decreasing to increasing (+ to - or - to +), it must first pass through a function value of zero. It is impossible to go from a negative number to a positive number without first passing through zero on the number line. So, if we find the places where the first derivative is equal to zero, we can examine the numbers between the zero values to find if they are positive or negative. So, obviously, we must first find where f'(x)=0:

3x2-3=0

3x2=3

x2=1

So, x=-1 or x=1

f'(-1)=0 and f'(1)=0, so examine the values between -1 and 1 to find what is happening to the slopes between these zero-slope locations. Also, you can examine the x-values greater than 1 and less than -1 to see what is happening on those intervals as well. Since the slopes would have to equal zero before they change from increasing to negative, and since we found that only two x-values give zero slope, whatever the value of the slope is found to be on one interval can be generalized to be the slope for any number on that interval.

f'(-2)=3(-2)2-3=(3)(4)-3=12-3=9

Since x=-2 is less than x=-1, and f'(-2)=24, we can say that any number less than -1 will have a positive slope, so the interval from negative infinity to -1 on f(x) is increasing.

x=0 is between x=-1 and x=1.

f'(0)=3(0)2-3=(3)(0)-3=0-3=-3

Since this value between x=-1 and x=1 results in a negative value for f'(x), then the function f(x) is decreasing between x=-1 to x=1.

x=2 is greater than x=1.

f'(2)=3(2)2-3=(3)(4)-3=12-3=9

Since this value greater than x=1 results in a positive value for f'(x), then the function f(x) is increasing for all values greater than x=1.

Therefore, in the end, we get that this function is increasing on the intervals (-infinity,-1) and (1,infinity).