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Given that the definition domain of function f (x) = 2asin (2x + π / 3) + B is [- π / 6, π / 6], the maximum value of function is 2 and the minimum value is 0, find the value of a and B
a=1
b=0
It is known that the definition domain of the function f (x) = 2asin (2x - π / 3) + B is [0, π / 2]. The maximum value of the function is 1 and the minimum value is 1
5, find the value of a and B
One step is from - π / 3 ≤ 2x - π / 3 ≤ 2 π / 3. We get - √ 3 / 2 ≤ sin (2x - π / 3) ≤ 1,
The domain of definition is [0, π / 2]
So 2x belongs to (0, π)
-π/3
It is known that the definition domain of the function f (x) = 2asin (2x - π / 6) + B (a > 0) is [0, π / 2], the maximum value of the function is 1, and the minimum value is 5. Find the values of a and B
When sin (2x π / 6) = 1,
f(x)|max＝2a+b＝1 ①;
When sin (2x - π / 6) = - 1,
f(x)|min＝-2a+b＝-5.②.
The results of solution 1 and 2 show that a = 3 / 2, B = - 2
How to find the maximum and minimum of the function f (x) = 2x-1, x + 1, X ∈ [3.5]?
Derivation of functions
f'(x)=2+1/x2
In the interval x ∈ [3.5], f '(x) > 0, so f (x) is an increasing function
So the maximum value of F (x) is f (5) = 54 / 5, f (3) = 20 / 3
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