Y = x-ln (1 + x) inflection point and concave convex interval

Y = x-ln (1 + x) inflection point and concave convex interval

y=X-Ln(1+X）
∴y'=1-1/(x+1)
∴y''=(x+1)^(-2)
Let y '' = 0
∴x=-1
But x = - 1 is not in the domain of function definition
There is no turning point
"Y 0" is established
Only concave interval (- 1, + ∞)

Find the concave convex interval and inflection point y 'denominator is 1 + X'2 y' = (1 + X '2)'2 why?

Because ln't = 1 / T = T ^ - 1
(t^-1)‘=-t^-2=-1/t^2
y‘ =ln‘ （1+x'2）=2x/（1+x'2）
y’’=[2x（1+x‘2)^-1]‘
=2（1+x‘2)^-1-2x（1+x‘2)^-2*2x
=2(1-x‘2)/(1+x‘2)^2

For the function y = (1 / 2) ^ (x ^ 2-6x + 17), find its monotone interval

This problem only requires the monotone interval of its index!
The exponent is converted to (x-3) ^ 2 + 8
If the base number is 1 / 2, it means that y is a decreasing function, and that the monotonicity of the function should be opposite to the function above the exponent
Then the increasing interval of the original function is (- infinite, 3), and the decreasing interval of the original function is (3, infinite),

Find the monotone interval of the function y = Log1 / 2 [(1-x) (x + 3)]!

Solution: the function y = Log1 / 2 x is a minus function
Therefore, the decreasing interval of the original function is also equal to the increasing interval of the function y = (1-x) (x + 3)
The increasing interval of the function y = (1-x) (x + 3) is (- ∞, - 1)
Because of the definition of the original function or (1-x) (x + 3) > 0, the solution is - 3

It is proved that the function y = x + X is a monotone decreasing function on the interval (0,1)

Let any x1, X2 ∈ (0,1], and x1, then f (x1) - f (x2) = X1 + 1 / x1-x2-1 / x2
=(x1-x2)+(x2-x1)/x1x2
=(x1-x2)+[(x2-x1)]/x1x2
=(x1-x2)[1-1/x1x2]
=(x1-x2)[(x1x2-1)/(x1x2)]
x1-x2<0,x1x2>0,
X1, X2 ∈ (0,1], then x1x2 < 1
∴(x1-x2)[(x1x2-1)/(x1x2)]>0.
So f (x1) > F (x2)
So f (x) is a minus function on (0,1]

The monotone decreasing interval of the function y = 1 / 1 of X-1 is?

(negative infinity, 1) monotone decreasing
(1, positive infinity) monotone decreasing

Finding the monotone interval of function y equal to 1 / 1 of X + 1

First, we can find the monotone interval of Y1 = x + 1, then we can get the monotone interval of y = 1 / (x + 1) which is opposite to Y1. Note that x in y = 1 / (x + 1) is not equal to - 1
The monotone increasing interval of Y1 is r
So y = 1 / (x + 1) monotone decreasing interval is (- infinity, - 1) U (- 1, + infinity)

How to find the monotone increasing interval of function y = 1-x without drawing pictures

Let x1.x2 be any two numbers in the domain of y = 1 / (1-x), and X1 < x2 means that f (x1) - f (x2) = 1 / (1-x1) - 1 / (1-x2) = (1-x2) / (1-x2) (1-x1) = (x1-x2) / (1-x2) (1-x1) = (x1-x2) / (1-x2) (1-x1) if X1 and X2 belong to (1, positive infinity), i.e., 1-x2 < 0,1

It is proved that the function y = 1 / 1 of X-1 is a monotone decreasing function on the interval (1, positive infinity)

Let x1, X2 ∈ (1, + ∞), and x1 ＜ X2, so: 1 ＜ x1 ＜ x2
Therefore, x1-1 > 0, x2-1 > 0, x2-x1 > 0
Because f (x1) - f (x2) = 1 / (x1-1) - 1 / (x2-1) = (x2-x1) / [(x1-1) (x2-1)] > 0
Therefore, the function y = 1 / (x-1) is a monotone decreasing function on the interval (1, + ∞)

How to find the monotone decreasing interval of function y = (1-x) / (1 + x)

y=(-1-x+2)/(1+x)
=(-1-x)/(1+x)+2/(1+x)
=-1+2/(1+x)
So X-1 decreases
The minus interval is (- ∞, - 1) and (- 1, + ∞)