It is known that the function f (x) is an odd function defined on R. when x > 0, f (x) = - 2x & # 178; + 3x + 1, and x < 0, Finding the analytic expression of F (x)

It is known that the function f (x) is an odd function defined on R. when x > 0, f (x) = - 2x & # 178; + 3x + 1, and x < 0, Finding the analytic expression of F (x)

If x ＜ 0, then - x ＞ 0, f (- x) = - 2 (- x) & # 178; + 3 (- x) + 1 = - 2x # 178; - 3x + 1, and because f (x) is an odd function in the domain R, then - F (x) = - 2x # 178; - 3x + 1, then f (x) = 2x # 178; + 3x-1, so when x ＜ 0, the analytic expression of F (x) is f (x) = 2x # 178; + 3x-1

1. If f (x) = 3x & # 178; - 5x + 2, then f (a + 3)=____ 2. F (x) = 2x & # 178; + 1, find the analytic expression of F (x + 2)
Given the inequality (1 / 3) ^ X & # 178; - 8 > 3 ^ - 2x about X, then the solution set of the inequality is__________

f(a+3)
=3(a+3)²-5(a+3)+2
=3a²+13a+14
f(x+2)
=2(x+2)²+1
=2x²+8x+9

Given f (x + 1) = x & # 178; + 3x + 1, find the analytic expression of F (x)
It says: let x + 1 = u, then x = U-1
Substituting knowledge: F (U) = (U-1) &# 178; + 3 (U-1) + 1
Calculate U & # 178; + U-1
But at this time, he said that because the number of the function has nothing to do with the letter, f (x) = x & # 178; + X-1
But it clearly says u = x + 1. Shouldn't it be counted in U + 178; + U-1 instead of u = x + 1

No, because it's f (U), not f (x). The function expression has nothing to do with the expression of the independent variable, but only with the corresponding rule. Here, find f (x) and write it as f (x) = x & # 178; + X-1. In fact, it's the same as f (U) = u & # 178; + U-1

F (2x + 1) = 3x-5, find f (x) =?

f(2x+1)=3x-5=3(2x+1)/2-13/2 f(x)=3x/2-13/2=(3x-13)/2

F (3x + 4 / 2x-1) = x + 5 for y = f (x)

Let t = (3x + 4) / (2x-1), then: 2tx-t = 3x + 4, (2t-3) x = t + 4, ｜ x = (T + 4) / (2t-3), ｜ f (T) = (T + 4) / (2t-3) + 5. That is: f (x) = (x + 4) / (2x-3) + 5. [x ≠ 3 / 2]

F (x) = 2x square + 3x + C and f (1) = 5, C =?
Process thank you

F (1) = the square of 2x1 + 3x1 + C = 5
2+3+c=5
c=0

√(3x-2/x-1)-5=√(2x+5)
----------- --------
√(3x-2/x-1) -5 = √(2x+5)
Wrong title: [√ (3x-2)] - 5 = [√ (2x + 5)]

√(3x-2) - 5 = √(2x+5)
Simultaneous square
(3x-2) - 10√(3x-2) + 25 = 2x+5
x+18 = 10√(3x-2)
x² + 36x + 324 = 100(3x-2)
x² - 164x + 524 = 0
Then it can be solved by the formula method of quadratic equation of one variable

f （x）＝a/3x^3＋b/2x^2－a^2x（a＞0）
Let X1 and X2 be the two extreme points of F (x), | x1 | + | x2 | = 2, (1) prove that 0 ＜ a ＜ = 1, (2) find the maximum value of B

F (x) = (A / 3) x ^ 3 + (B / 2) x ^ 2 - (a ^ 2) x (a > 0) - right?
f '（x）＝ax^2＋bx－(a^2)
From ax ^ 2 + BX - (a ^ 2) = 0, we can see that the equation △ = B ^ 2 + 4A ^ 3 > 0 has two solutions,
In this case, X1 + x2 = - B / A, X1 * x2 = - (a ^ 2) / a = - A. if you know the different sign of X1 and X2, you might as well set x1

Find the range of F (x) = 5-x ＋ 3x-1

Let t = √ (3x-1)
There are: x = (T ^ 2 + 1) / 3 (t ≥ 0)
Then: F (x) = 5 - (T ^ 2 + 1) / 3 + T (t ≥ 0)
===> f(x)＝(－t^2 +3t +14)/3 （t ≥ 0)
===> f(x)＝(－t^2 +3t +14)/3 （t ≥ 0)
===> f(x)＝[65/4－(t－3/2)^2 ] / 3 （t ≥ 0)
===> f(x) ≤ 65/12

Find the following function range (expressed by interval) (1) y = x & sup2; - 3x + 4 (2) f (x) = 2x & # 178; - 2x + 4 under the root sign

Y = (x-3 / 2) 2 + 7 / 4, so the interval should be closed from 7 / 4 to infinity
F (x) = 2 (x-1 / 2) 2 + 7 / 2 under the root sign, so the interval should be closed from 7 / 2 under the root sign to positive infinity