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    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
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    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.
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    (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
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    (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 f'(x) > 0 at all points. This must be shown to be the case. Clearly, we have f'(x)=3x^2+1. Note that this is just a +ve parabola with a minimum value of 1. This means that f'(x) \geq 1 > 0 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 f'(x) to be 0. All you have to do is to show that f'(x) is either >0 or <0 using the same logic as above.
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    (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.
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    (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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