The range of function y = - X & # 178; + 2x + 3 (0 ≤ x ≤ 3) is

The range of function y = - X & # 178; + 2x + 3 (0 ≤ x ≤ 3) is

According to the characteristics of the function, it is decomposed into factors, y = - (x-3) (x + 1). It is found that the intersection of the function and the X axis is - 1 and 3, and the opening is downward
Using the matching method, it is found that its symmetry axis is x = 1, the intersection of the image and the symmetry axis is (1,4), and the intersection of the image and the Y axis is (0,3)
According to the definition field, we can see that the value range is [0,4]
Let the derivative of function f (x) be f '(x), and f (x) = x ^ 2 + 2x * f' (1), then f '(0) is equal to?
f(x)=x²+2x*f'(1)
Here f '(1) is a constant, that is, the coefficient of X is 2F' (1)
Then f '(x) = 2x + 2 * f' (1)
Let x = 1
f'(1)=2+2*f'(1)
So f '(1) = - 2
So f '(x) = 2X-4
So f '(0) = - 4
It is proved that cot2 α = (1 + SiN4 α + Cos4 α) / (1 + SiN4 α - Cos4 α)
(1+sin4α+cos4α)/(1+sin4α-cos4α)
=(2cos^2(2α)+2cos2αsin2α)/(2sin^2(2α)1+2cos2αsin2α)
=cos2α/sin2α=cot2α
Let log (x / a) = X-1
Zero
The function is a decreasing function
loga（1-a/x）>1
loga（1-a/x）>loga(a)
So 0
It is proved that when θ is an acute angle, 1
sinθ+cosθ=√2sin(θ+π/4)
∵0＜θ＜π/2
∴π/4＜θ+π/4＜3π/4
√2 /2＜sin(θ+π/4)≤1
∴1＜√2sin(θ+π/4)≤√2
That is: 1 ＜ sin θ + cos θ ≤ √ 2
If f (x) is an odd function defined on (- 1,1) and is a decreasing function, the inequality f (1-A) + F (1-A ^ 2) < 0 is solved
f(1-a)+f(1-a^2)>0
f(1-a)>-f(1-a^2)
F (x) is an odd function, so
f(1-a)>f(a^2-1)
Y = f (x) is defined on (- 1,1), so
-1
α. If β is acute angle and α + β > π / 2, then cos α is less than sin α
Ah, the proof is that cos α is less than sin β
If there is a problem in the title, it should be proved that cos α is less than sin β. Otherwise, a counterexample can be given. α = 30 ° and β = 70 ° satisfy the condition, but do not satisfy the final result. Cos 30 ° should be greater than sin 30 °
prove:
∵α+β>π/2,∴α>π/2-β
Both α and β are acute angles,
π / 2 - β is an acute angle
cosα
Prove that cos α is less than sin β?
－－－－－－－－
α+β>π/2
α>π/2-β
α and π / 2 - β are between 0 and π / 2, so cos α < cos (π / 2 - β) = sin β
α+β>π/2
β>π/2-α
α. β is acute angle
When 0
If f (x) is a decreasing function defined on (- 1,1), solve the inequality f (1-A) - f (a ^ 2-1)
F (x) is a decreasing function defined on (- 1,1)
f(1-a)-f(a^2-1)
f(1-a)-f(a^2-1)
If θ is an acute angle, the trigonometric function is used to prove that 1 < sin θ + cos θ < π / 2
If θ is an acute angle, the trigonometric function [line] is used to prove that 1 ＜ sin θ + cos θ ＜ π / 2. Wrong number
Sin θ + cos θ = radical 2 * sin (θ + π / 4)
Zero
If f (x) is a decreasing function on R and the image of F (x) passes through points a (0,3) and B (3, - 1), then the solution set of the inequality | f (x + 1) - 1 | 2 is______ .
From | f (x + 1) - 1 | 2, we get - 2 | f (x + 1) - 1 | 2, that is - 1 | f (x + 1) | 3. Because f (x) is a decreasing function on R, and the image of F (x) passes through points a (0, 3), B (3, - 1), so f (3) | f (x + 1) | f (0). So 0 | x + 1 | 3, - 1 | x | 2