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F (x) = | X-2 |, what is the derivative at point x = 2, or does it not exist?
Left derivative - 1, right reciprocal 1, they are not equal, so they do not exist
The image of the function f (x) = cos2x-sin2x + 2 is shifted m units (M > 0) to the left along the x-axis, and the image of the function is symmetric about the line x = 17 π / 8,
Then the minimum value of M is?
Would you please give me the specific steps
I see
It's no use for me to score
Thank you!!!!!
The original formula = (1-cos2x) / 2-sin2x + 3 · (1 + cos2x) / 2 = 1 / 2-cos2x / 2-sin2x + 3 / 2 + 3cos2x / 2 = cos2x-sin2x + 2 put forward a root sign 2 -- don't tell me you don't know = root sign 2 × (root sign 2 / 2 × cos2x - root sign 2 / 2 × sin2x) and because sin45 ° and cos45 ° are equal to half root
This paper discusses whether the function is differentiable or continuous at the point x = 1, if f (x) = sin (x-1) / (x-1) x is not equal to 1, if 0 x = 1
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