Given the function y = 2x & # 178; + 3x-k, if y is greater than 0 for any x, then the value range of K is? Seeking process!

Given the function y = 2x & # 178; + 3x-k, if y is greater than 0 for any x, then the value range of K is? Seeking process!

Because for any x, Y > 0
So Delta
The equation 2F ^ 2 (x) + 2BF (x) + 1 = 0 of the function f (x) with the domain r has eight different real roots
F (x) = {lgx |, x > 0; - x ^ 2-2x, X
If we draw the image of F (x), we can see that when 00
h(1)>0
Zero
Given the function y = 2x & # 178; + 3x-k, if y is greater than 0 for any x, then the value range of K is____
I remember that there seem to be two cases: when △ 0, when △ 0, a closed interval?
k＞-9/8
Analytic results show that when △ 0, y is always less than 0 because y = 2x & # 178; + 3x-k has no focus with x-axis and Y is always less than 0; △ 0, y ≤ 0; if and only if △ 0, y can be greater than 0
Given the function f (x) = SiNx + 1 / SiNx, find its range
f(x)=sinx+1/sinx
Because SiNx ∈ [- 1,1]
And y = x + 1 / X is a double hook function
Is a decreasing function on [- 1,0), (0,1]
So f (x) ≤ - 1 + 1 / (- 1) = - 2 or F (x) ≥ 1 + 1 / 1 = 2
So the range of the function is (- ∞, - 2] ∪ [2, + ∞)
f(x)=sinx+1/sinx
When SiNx > 0
f(x)=sinx+1/sinx≥2
If and only if SiNx = 1, the equal sign holds
sinx
Given f (3x-4) = 2x & # 178; - 1, find f (5), f (x)
F (x) = 2 [(x + 4) / 3] & # 178; - 1
F (5) = brought into the above formula
F (x) = 2 [(x + 4) / 3] & # 178; - 1
Can you be more detailed
The range of function f (x) = √ (1 / 2-sinx) is
sinx∈[-1,1]
Then 1 / 2-sinx ∈ [- 0.5,1.5]
Then f (x) ∈ [0, √ 1.5]
-1
Given f (2x) = 3x & # 178; - x + 1, find f (x)
f(2x)=3x²-x+1
Let x = x / 2
f(2*x/2) = 3(x/2)^2-(x/2)+1
f(x) = 3/4 x^2 - 1/2 x + 1
Given f (2x) = 3x & # 178; - x + 1, find f (x).
x'=2x
x=x'/2
f(x)=3(x/2)²-x/2+1
Divide each item on the right into a number multiplied by 2x or the square of 2x, and then replace 2x on both sides with X
It is known that the domain of definition of the function y = Log1 / 2 (x ^ 2 + ax + A-3 / 4) is R. the value range of a is obtained
A:
Y = Log1 / 2 (x ^ 2 + ax + A-3 / 4) the domain is r
Then the true number f (x) = x ^ 2 + ax + A-3 / 4 > 0 is constant on R
So: discriminant = a ^ 2-4 (A-3 / 4)
Let f (x) = x & # 179; - 2x & # 178; + 3x + 3, find f (0), f (1), f (- 1), f (x + 1)
f（0）= 0 ³ - 2 × 0 ² + 3 × 0 + 3 = 3 f（1）= 1 ³ - 2 × 1 ² + 3 × 1 + 3 = 5 f（- 1）= （- 1）³ - 2 × （- 1）² + 3 × （- 1）+ 3 = - 3 f（x + 1）= ...
solution
f(0)=0³-2×0²+3×0+3=3
f(-1)=(-1)³-2×(-1)²+3×(-1)+3
=-1-2-3+3
=-3
f(1)=1³-2×1²+3×1+3
=1-2+3+3
=5
f(x+1)=(x+1)³-2(x+1)²+3(x+1)+3
=x³+3x²+3x+1-2(x²+2x+1)+3x+6
=x³+x²+2x+5
If the domain of the function y = Log1 / 2 (AX ^ 2 + AX = 1) is all real numbers, then the value range of a?
Correction: 'ax ^ 2 + AX = 1' should be 'ax ^ 2 + ax + 1'
Because the domain is all real numbers
So when x takes all real numbers, the value range of ax ^ 2 + ax + 1 is > 0
The equivalent function y = ax ^ 2 + ax + 1 is Y > 0
1. When a = 0, y = 1 > 0
2. When a ≠ 0, y = ax ^ 2 + ax + 1 is a quadratic function
At this time, in order to meet the requirements, a > 0 (opening upward) and Δ = a ^ 2-4a > 0
The solution is a > 4
So the value range of a is a = 0 or a > 4