Given the function f (x) = / x ^ 2-4x-5 /, if in the interval [- 1,5], the image of y = KX + 3K is above the function f (x), the value range of K is obtained

Given the function f (x) = / x ^ 2-4x-5 /, if in the interval [- 1,5], the image of y = KX + 3K is above the function f (x), the value range of K is obtained

∵ f (x) = | X & # 178; - 4x-5 | = | (x + 1) (X-5) | in the interval [- 1,5],
Then f (x) = - (x + 1) (X-5) = - X & # 178; + 4x + 5, - 1 ≤ x ≤ 5
X-axis = - B / 2A = 2, vertex is (2,9)
And y = KX + 3K = K (x + 3) is a straight line passing through the point (- 3,0)
In order to satisfy the condition that the image of y = KX + 3K is above the function f (x), the critical condition between the line and f (x) must be found
Let f (x) and the line y = KX + 3K have only one intersection at - 1 ≤ x ≤ 5, and let f (x) be the tangent point (m, n)
The derivative of F (x) is y ′ = - 2x + 4 = K
Let K be taken into y = K (x + 3)
y=(-2x+4)(x+3)=-2x²-2x+12
Then n = - M & # 178; + 4m + 5
n=-2m²-2m+12
The solution of the above two equations is m = 1 (M = - 7)
Here k = 2
So the range of K is k > 2

When k > 2, why is the image of y = KX + 3K above the function f (x) = - x2 + 4x + 5 in the interval [- 1,5]?
Please answer in detail to the last step,

Let g (x) = KX + 3k-f (x) = KX + 3K + X & sup2; - 4x-5 = x & sup2; + (K-4) x + 3k-5
If in the interval [- 1,5], the image of y = KX + 3K is above the image of function f (x) = - X & sup2; + 4x + 5
Then in the interval [- 1,5], G (x) is always greater than 0
That is g (x) = x & sup2; + (K-4) x + 3k-5 > 0
When - (K-4) / 26,
The minimum value of G (x) is g (- 1) = 1-k + 4 + 3k-5 = 2K > 0
When k > 6, G (x) ≥ g (- 1) > 0
When - 1 ≤ - (K-4) / 22, in the interval [- 1,5], G (x) > 0
On the interval [- 1,5], the image of y = KX + 3K is above the image of F (x) = - X & sup2; + 4x + 5

If f (x) = (K's Square - 3K + 2) x's square + 2x + m + 1 is an odd function, then K=__ m__

f(x)=(k²-3k+2)x²+2x+m+1
f(-x)=(k²-3k+2)x²-2x+m+1
-f(x)=-(k²-3k+2)x²-2x-m-1
(k²-3k+2)x²-2x+m+1=-(k²-3k+2)x²-2x-m-1
(k²-3k+2)x²+m+1=-(k²-3k+2)x²-m-1
m+1=-m-1
m=-1
k²-3k+2=-k²+3k-2
k²-3k+2=0
(k-1)(k-2)=0
k=1 k=2

F (x) = (k ^ 2-3K + 2) x ^ 2 + 2x + m + 1 is an odd function
Then k =?
m=?

K is 3 / 2, M is - 1, f (- x) = - f (x), so x ^ 2 is 0, M-1 = 0