Find the maximum and minimum of
f:[-3π/4, 5π/4] --> R, f(x) = asin(x-b)+c
where a, b and c are positive constants
thanks in advance : ))
edit: and this one too:
find the value(s) of x for which |2x+1| - |3-x|=9
TY!!
For the maximum minimum question, there are many ways to do it, but we will try the f'(x) = 0 way:
f(x) = asin(x-b)+c
f'(x) = acos(x-b)
Let f'(x) = 0
Now the original domain is
, but you have to alter this by subtracting 'b' from it, as done with the actual expression of f(x) (x-b)
So new domain is:
acos(x-b) = 0
cos(x-b) = 0
But now make sure that these two solutions lie within the domain above, as 'b' is a positive constant, then a negative number minus a negative number will always yield a smaller, negative number, so the negative pi/2 solution is valid, and the positive pi/2 solution is unknown because it would depend on the value of 'b', that is, how large the constant of 'b' actually is. So do they give us more information about the constants? For this example, I would just assume that they are both valid solutions because theres no further information given.
Sub this into f(x) to get the exact y-value of the max/min:
 = a sin(\frac{\pi}{2}) + c = a+c)
 = a sin(-\frac{\pi}{2}) + c = c-a)
Now that was the extremely long way; the shorter way can be done in three lines:
From transformations, you know the amplitude will usually give you the maximum and minimum with circular functions, so in this case it would be
but because there is a vertical translation, then the max and min both move up by 'c' units, so our new max and min values would be: a+c and -a+c.
I hope this helped you
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