Mathematical problem f (x + 1) = 2x ^ 2 + 1, then f (x)=

Mathematical problem f (x + 1) = 2x ^ 2 + 1, then f (x)=

f(x+1)=2x^2+1
Let t = x + 1
x=t-1
f(t)=2(t-1)²+1
=2t²-4t+3
So:
f(x)=2x²-4x+3

If f (x) = sin ^ 2x-1, the period is?
Let's see what's wrong with my calculation. It's different from your process

Because cos2x = 1-2sin & # 178; X, so Sin & # 178; X = (1-cos2x) / 2
f(x)=(1-cos2x)/2-1=-1/2*cos2x-1/2.
The period is 2 π / 2 = π

The solution of 8x-3x-2105

8x-3x=105
5x=105
x=21
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A simple calculation method: 0.25 × 1 / 7 + 1 / 4 × 6 / 7-1 / 4

0.25×1/7+1/4×6/7-1/4
=0.25x1/7+0.25x6/7-0.25
=(1/7+6/7-1)x0.25
=0x0.25
=0
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If | X-5 | + X + 1 is a constant, then X_______ This constant is equal to______

If | X-5 | + X + 1 is a constant, then [x ≤ 5], this constant is [6]

Divide the number a by 13 to make 7, and the number B by 13 to make 9. Now multiply the number a and the number B, and divide the product by 13 to make 9______ ．

Let a be 13A + 7, and B be 13b + 9, then: a × B = (13a + 7) (13b + 9) = 13 × 13ab + 13 × 9A + 13 × 7b + 63, we can see that the first three terms can be divided by 13, so the remainder of the product divided by 13 is the remainder of 63 divided by 13, and 63 △ 13 = 4 So the answer is: 11

How to divide 9 by 0.125

0.125=1/8
9 divided by 0.125 = 9 / (1 / 8) = 9 * 8 = 72

How to solve hyperbolic equation with known asymptote equation?

It is known that the asymptotic equation is ax + by = 0
Then let the hyperbolic equation be a ^ 2x ^ 2-B ^ 2Y ^ 2 = K
Then substitute a coordinate to get K

Given f (x) = {1, X is greater than or equal to 0 - 1, X is less than 0, then the solution set of inequality x + (x + 2) * f (x + 2) less than or equal to 5 is?

F (x) = {1, X is greater than or equal to 0 - 1, X is less than 0}
x+(x+2)*f(x+2)《5
Discussion by situation
When x + 2 ″ 0, i.e. x ″ - 2
f(x+2)=1
X + (x + 2) * f (x + 2) 5 into
x+(x+2）《5
2x《3
x《3/2
So it's - 2 "X" 3 / 2
x+2

How much is 1 / 4 times 99 plus 1 / 4

1/4×99＋1/4
=1/4(99+1)
=1/4×100
＝25
or
＝99/4+1/4=100/4=25