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The maximum gradient of the curve y=x^4 -4x^3 + 4x^2 +2

in the range 0<= x <= 2.2

I did it by working out the 2nd and 3rd derivative, however in the answer (which shows the full working) it says 'the gradient will attain its maximum at either at its endpoint or its maximum turning point (i think it means maximum turning point of the first derivative)'

and the maximum gradient turns out to be when x=2.2, wondering why I have to consider the values at the beginning and end of the intervals

in the range 0<= x <= 2.2

I did it by working out the 2nd and 3rd derivative, however in the answer (which shows the full working) it says 'the gradient will attain its maximum at either at its endpoint or its maximum turning point (i think it means maximum turning point of the first derivative)'

and the maximum gradient turns out to be when x=2.2, wondering why I have to consider the values at the beginning and end of the intervals

If you are given an interval then there may not be a turning point in that interval. If there is no turning point, then the maximum has to be either at the start or the end of the interval. Think of, for example, the maximum value of $f(x) = x$ in the domain $0 < x < 2$, or the maximum value of $f(x) = -x$ in the same domain. Even if your function has turning points outside of the interval, if it doesn't have any turning points inside the interval then it will be similar to these two examples where the maximum/minimum are at the endpoints.

yep, i understand what you mean, but

i clearly worked out a turning point (by turning point i mean the second derivative = 0, and using third derivative to make sure it was a maximum) at x = 1-1/root3, but it isnt and instead is when x =2.2, i dont really see why that should be the case, if there clearly is a turning point in the interval.

i clearly worked out a turning point (by turning point i mean the second derivative = 0, and using third derivative to make sure it was a maximum) at x = 1-1/root3, but it isnt and instead is when x =2.2, i dont really see why that should be the case, if there clearly is a turning point in the interval.

cvat

yep, i understand what you mean, but

i clearly worked out a turning point (by turning point i mean the second derivative = 0, and using third derivative to make sure it was a maximum) at x = 1-1/root3, but it isnt and instead is when x =2.2, i dont really see why that should be the case, if there clearly is a turning point in the interval.

i clearly worked out a turning point (by turning point i mean the second derivative = 0, and using third derivative to make sure it was a maximum) at x = 1-1/root3, but it isnt and instead is when x =2.2, i dont really see why that should be the case, if there clearly is a turning point in the interval.

Because a turning point is not necessarily a maximum, it's just a local maximum. If the interval is big enough, the graph may well turn again and climb higher then the first turning point. Check the y value at your calculated turning point and then the y value at x=2.2. Also, look at some cubic polynomial graphs to see what I mean.

Original post by T King

The question is asking for the steepest gradient NOT for a turning point

Please think twice before reviving 14 year old threads.

Original post by T King

My student came to me with this exact question today!!!! I know they use student room so thought it would be helpful if I corrected the wrong train of thought given here!

Fair point.

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