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Given that f (x) is a quadratic function, f '(x) is its derivative function, and f' (x) = f (x + 1) + X2 is constant for any x ∈ R, f '(x) = f (x + 1) + X2, the analytical expression of F (x) is obtained
Let f (x) = AX2 + BX + C (where a ≠ 0),
Then f '(x) = 2aX + B,
∵f（x+1）=a（x+1）2+b（x+1）+c=ax2+（2a+b）x+a+b+c．
From the known, 2aX + B = (a + 1) x2 + (2a + b) x + A + B + C,
∴
a+1＝0
2a+b＝2a
a+b+c＝b ， The solution is a = - 1, B = 0, C = 1,
∴f（x）=-x2+1．
It is known that the image with quadratic function y = f (x) passes through the coordinate origin, and its derivative is f '(x) = 6x-2 The sum of the first n terms of the sequence {an} is Sn, and the points (n, Sn) (n belongs to n) are on the image of the function y = f (x) (1) General formula for finding sequence {an} (2) Let BN = 2 / ana (n + 1), TN is the sum of the first n terms of the sequence {BN}, and find the minimum positive integer m such that TN < m / 20 belongs to n * for all n
Didn't you learn points in high school?
① The quadratic function crosses the origin and is set as f (x) = ax ²+ bx
f'(x)=2ax+b=6x-2
So a = 3, B = - 2
f(x)=3x ²- 2x
Sn=3n ²- 2n
S(n+1)=3(n+1) ²- 2(n+1)
a(n+1)=S(n+1)-Sn=6n+1
So an = 6n-5
②bn=2/ana(n+1)=2/(6n-5)(6n+1)=[1/(6n-5)-1/(6n+1)]/3
∴Tn=[1-1/7+1/7-1/13+1/13-1/19+...+1/(6n-5)-1/(6n+1)]/3
=[1-1/(6n+1)]/320[1-1/(6n+1)]/3
For any n ∈ n *, there is 20 [1-1 / (6N + 1)] / 320 / 3
So m = 7
It is known that the image with quadratic function y = f (x) passes through the coordinate origin, and its derivative is f '(x) = 2x + 2. Sequence {an It is known that the image with quadratic function y = f (x) passes through the coordinate origin, and its derivative is f '(x) = 2x + 2 The sum of the first n terms of the sequence {an} is Sn, and the points (n, Sn) (n belongs to n) are on the image of the function y = f (x). (1) General formula for finding sequence {an} (2) Let BN = 2 / ana (n + 1), TN is the sum of the first n terms of the sequence {BN}, and find the expression of TN
(1) The derivative of quadratic function y = f (x) is f '(x) = 2x + 2
The quadratic function f (x) = x can be set ²+ 2x+C
The image of the quadratic function y = f (x) passes through the coordinate origin
‡ C = 0, i.e. quadratic function f (x) = X ²+ 2x
The sum of the first n terms of ∵ sequence {an} is Sn, and the points (n, Sn) (n belongs to n) are on the image of function y = f (x)
∴Sn=n ²+ 2n
∴a1=S1=3
an=Sn-S(n-1)
=n ²+ 2n-[(n-1) ²+ 2(n-1)]
=2n+1
(2)、∵bn=2/ana(n+1)
=2/[(2n+1)(2n+3)]
=1/(2n+1)-1/(2n+3)
∴Tn=b1+b2+b3+……+bn
=2/a1a2+2/a2a3+3/a3a4+……+2/ana(n+1)
=(1/3-1/5)+(1/5-1/7)+……+[1/(2n+1)-1/(2n+3)]
=1/3-1/(2n+3)
=2n/[3(2n+3)]
It is known that the image of quadratic function y = FX passes through the origin, its derivative function f'x = 6x-2, the primary function is y = GX, and the solution set of inequality GX > FX is {x| 1 / 3
From the derivative function, we can know that y = FX = 3x square - 2x + T, and then substitute t = 0.y = FX = 3x square - 2x through the origin (0,0)
Let y = GX = ax + B, and the solution set of GX > FX be {x | 1 / 3
Given that the image of the quadratic function y = f (x) crosses the coordinate origin, and 1 ≤ f (- 1) ≤ 2,3 ≤ f (1) ≤ 4, find the value range of F (- 2) Undetermined coefficient method When to use the undetermined coefficient method thank
solution
Let the quadratic function be f (x) = ax ²+ bx
f(1)=a+b
f(-1)=a-b
f(-2)=4a-2b=Af(1)+Bf(-1)
That is, 4a-2b = a (a + b) + B (a-b) = (a + b) a + (a-b) B
A+B=4
A-B=-2
So a = 1, B = 3
That is, f (- 2) = f (1) + 3f (- 1)
1≤f(-1)≤2,3 ≤ 3f(-1)≤ 6 (1)
3≤f(1)≤4,(2)
(1)+(2)
6≤f(1)+3f(-1)≤10
That is, 6 ≤ f (2) ≤ 10
It is known that the image with quadratic function y = f (x) passes through the coordinate origin, its derivative function is f '(x) = 6x-2, and the sum of the first n terms of the sequence {an} is SN The points (n, Sn) are all on the function y = f (x) image, and the general term formula of the sequence {an} is obtained
But the solution is an = 6n-5
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