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# Differentiation Functions HELP watch

1. Use differentiation to explain why f(x) = x^3+x is an increasing function for all real values of x.
I did
f'(x)= 3x^2+1
but then 3x^2>-1
and then x^2>-1/3
But the answer says x^2 can't be negative and so f'(x) is an increasing function for all real values of x. x^2>0
I'm a bit confused here

For Q2
Is the function f(x)=3-3x-x^3 an increasing or decreasing function?
f'(x)= -3x^2-3
so then I did -3x^2=-3
3x^2=-3
x^2=-1
but it this is also wrong
2. Your inequality that 3x^2 >-1 is incorrect. If you have done graph transformations, start from the graph x^2 and do the corresponding graph transformations to get to 3x^2+1 and you should see this is the case. You can start from the fact that x^2 >=0 for all real values of x. If you manipulate this inequality to get the LHS to be 3x^2+1 you should arrive at the correct answer. You can do a similar procedure for the second question.
3. (Original post by Pepperpeople)
Your inequality that 3x^2 >-1 is incorrect. If you have done graph transformations, start from the graph x^2 and do the corresponding graph transformations to get to 3x^2+1 and you should see this is the case. You can start from the fact that x^2 >=0 for all real values of x. If you manipulate this inequality to get the LHS to be 3x^2+1 you should arrive at the correct answer. You can do a similar procedure for the second question.
But why did I get negative 1 instead of a positive number for the 2nd one
4. (Original post by Pakora99)
Use differentiation to explain why f(x) = x^3+x is an increasing function for all real values of x.
I did
f'(x)= 3x^2+1
but then 3x^2>-1
and then x^2>-1/3
But the answer says x^2 can't be negative and so f'(x) is an increasing function for all real values of x. x^2>0
I'm a bit confused here
The idea is that if a function is increasing, then its derivative is at all points. This must be shown to be the case. Clearly, we have . Note that this is just a +ve parabola with a minimum value of 1. This means that as required.

For Q2
Is the function f(x)=3-3x-x^3 an increasing or decreasing function?
f'(x)= -3x^2-3
so then I did -3x^2=-3
3x^2=-3
x^2=-1
but it this is also wrong
Not sure why you are setting to be 0. All you have to do is to show that is either >0 or <0 using the same logic as above.
5. (Original post by Pakora99)
But why did I get negative 1 instead of a positive number for the 2nd one
You have attempted to solve in that equation whether f'(x) = 0 at any point which would show where a stationary point is. Since your equation there has no solutions that shows there is no stationary/turning point. For this question, you want to show whether f'(x) is always greater than or less than 0 as this will show whether the function is always increasing or decreasing respectively. You can use the procedure I described in my last post to do this.
6. (Original post by Pepperpeople)
You have attempted to solve in that equation whether f'(x) = 0 at any point which would show where a stationary point is. Since your equation there has no solutions that shows there is no stationary/turning point. For this question, you want to show whether f'(x) is always greater than or less than 0 as this will show whether the function is always increasing or decreasing respectively. You can use the procedure I described in my last post to do this.
Just realised x^2 cannot create a negative so x^2>=0 which means 3x^2+1 is an increasing function

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