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junior college 1 | H2 Maths
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Good evening! I need a bit of help with this question haha :)
Date Posted: 4 years ago
Views: 259
Good evening Candice! Looking at your question now.
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done
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clear
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I have done completing the square all the way.
For the last part, we see that f’(x), which equals to -2 [cos x/2 - 1]^2 - 1, has a greatest value of -1. This is because the lowest value of the squared expression is 0, and multiplying the squared expression by a negative number makes the greatest value zero.
So, f’(x) <= -1 < 0 for all x > 0.
And, when x = 0, f(x) = 8 sin 0 - sin 0 - 0 = 0.
Since f(x) = 0 and the gradient function f’(x) is always negative for all x > 0, f(x) < 0 for all real x > 0.
For the last part, we see that f’(x), which equals to -2 [cos x/2 - 1]^2 - 1, has a greatest value of -1. This is because the lowest value of the squared expression is 0, and multiplying the squared expression by a negative number makes the greatest value zero.
So, f’(x) <= -1 < 0 for all x > 0.
And, when x = 0, f(x) = 8 sin 0 - sin 0 - 0 = 0.
Since f(x) = 0 and the gradient function f’(x) is always negative for all x > 0, f(x) < 0 for all real x > 0.
Date Posted:
4 years ago
Good evening Candice! We need to use the gradient function to prove that f(x) < 0 for all x > 0, but it is not sufficient to conclude that f(x) < 0 if all f'(x) < 0, because the graph could well start on the positive side.
This is why we must also check the point when x = 0 to make sure it is not on the positive side.
This is why we must also check the point when x = 0 to make sure it is not on the positive side.
Thank you! :D
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