Given that the derivative of F (e ^ x) is 1 + X, then f (x)=

Given that the derivative of F (e ^ x) is 1 + X, then f (x)=

Because:
[f(e^x)]'=1+x
So:
f(e^x)=(1/2)x^2+x+c
Let e ^ x = t, so x = LNT
Namely:
f(t)=(1/2)ln^2t+lnt+c
So:
F (x) = (1 / 2) ln ^ 2x + LNX + C, C is constant
Let f (x) be an odd function defined on R, and when x > 0, f (x) = - 2x ^ 2 + 3x + 1, find the analytic expression and monotone interval of F (x)
Let x0, f (- x) = - 2 (- x) ^ 2 + 3 (- x) + 1 = - 2x ^ 2-3x + 1
That is - f (x) = - 2x ^ 2-3x + 1
∴f(x)=2x^2+3x-1
So f (x) = {- 2x ^ 2 + 3x + 1 x > 0
{0 x=0
{2x^2+3x-1 x
First of all, we can see that we need to use piecewise function to solve this problem. We know that x > 0. So when x0, we will replace X in the original formula with - x, f (- x) = - 2x ^ 2-3x + 1. Because it is an odd function, f (x) = - (- 2x ^ 2-3x + 1) = 2x ^ 2 + 3x-1. We can find the monotone interval. Just draw a sketch to find out the axis of symmetry, and remember to sum x > 0. And x0, We will replace X in the original formula with - x, f (- x) = - 2x ^ 2-3x + 1, because it is an odd function, so f (x) = - (- 2x ^ 2-3x + 1) = 2x ^ 2 + 3x-1, monotone interval can be found, just draw a sketch, find out the axis of symmetry, it is clear at a glance, remember to sum x > 0. And X
Given f (x) = x ^ 2 + the derivative f '(x) of ex-e ^ x, then f' (1)
y=x^2+ex-e^x
By using the derivative formula of sum difference of function, the following results are obtained
y'=(x^2)'+(ex)'-(e^x)'
=2x+e-e^x
Namely:
f(x)'=2x+e-e^x
f(1)'=2+e-e=2.
Using the derivative formula
f~（x）=2x+e-e^x
So f ~ (1) = 2
Let f (x) be an odd function defined on R, and if x > 0, f (x) = - 2x ^ 2 + 3x + 1
Try to find the analytic expression and monotone interval of function f (x)
X0.
f(-x)=-2x^2-3x+1=-f(x)=2x^2+3x-1
f(x) = -2x^2+3x+1 x>0
0 x=o
2x^2+3x-1 x
Derivative of e ^ X-1
RT
∵ the reciprocal of the x power of E or the x power of E, the reciprocal of 1 is 0, so the reciprocal above is the x power of E
The derivative of e ^ X-1 is e ^ X-1
It is known that f (x) is an odd function defined on R. when x is greater than 0, f (x) = - 2x square + 3x + 1, then f (- 2)=
After replacing x with - x, the analytic formula of F (- x) is - 2x square - 3x + 1. At this time, do you directly replace - 2 with - 2x square - 3x + 1 or F (- - 2))?
When x is greater than 0, f (x) = - 2x & # 178; + 3x + 1,
Take x0
∴f(-x)=-2(-x)²+3(-x)+1=-2x²-3x+1
∵ f (x) is an odd function
∴f(-x)=-f(x)
∴f(x)=-f(-x)=-(-2x²-3x+1)=2x²+3x-1
F (- 2) = 1
f(-2)=-f(2) f(2)=-1
So f (- 2) = 1
What is the derivative of F (x) = e ^ x-e ^ - x
f'(x)=e^x+e^(-x)
e^x+e^-x
Because the derivative of e ^ - x is - e ^ - X
This is the composite derivative, and - x also has to be derivative = - 1
Let f (x) be an odd function defined on R, and if x > 0, f (x) = - 2x * x + 3x + 1, try to find the analytic expression of F (x)
analysis
Since the function is odd, let - f (x) = f (x)
-x=x
So when x
How to find the derivative of F (x) = (A-X / A + x) e ^ x?
Predigestion, f (x) = [2A / (a + x) - 1] × e ^ x
Then, according to the multiplication derivation rule, f (x) = UV, f (x) derivative = u derivative × V + U × V derivative
F (x) derivative = - 2A / [(a + x) ^ 2] × e ^ x + [2A / (a + x) - 1] × e ^ x
={-2a/[(a+x)^2]+2a/(a+x)-1}×e^x
f(x)=g(x)h(x)
f'(x)=g'(x)h(x)+g(x)h'(x)
So according to your formula, the answer is: F '(x) = (- 1 / A + 1) e ^ x + (A-X / A + x) e ^ x = [A-1 / A + 1 + (1-1 / a) x] e ^ X
f（x）=（a-x/a+x）e^x
=[(a-x)/(a+x)]'*e^x+（a-x/a+x)*[e^x]'
=[2a/(a+x)^2+(a-x)/(a+x)]*e^x
=(a^2+2a-x^2)/(a+x)^2*e^x
It is known that f (x) is an odd function defined on R when x
Because of odd function, f (- x) = - f (x)
Because when X0
When x = 0, f (0) = ± 1
Finally, x > 0, x = 0, x = 0