Advanced Mathematics Help Draw a Circle Thank you 100

x->1-

f(x)-f(1))/x-1 = x^2+1-1^2-1/x-1 = (x+1) = 2

x->1+

f(x)-f(1)/x-1 = 3x-1-3*1+1/x-1 = 3

When approaching 1+, the value of 1- is different.

So it can't be guided.

f(x)(x->0+)(1+x)^1/2-(1-x)^1/2) = 0

f(x)(x->0-)ln(x+1) = 0

f(0) = ln1 = 0

So continuously. x->0+

f(x)-f(0)/x-0 = (1+x)^1/2-(1-x)^1/2/x = ((1+x)-(1-x))/x*((1+x)^1/2+(1-x)^1/2) = 2/((1+x)^1/2+(1-x)^1/2) = 1

x->0-

f(x)-f(0)/x-0 = ln(1+x)-ln(1+0)/x-0 = ln(1+x)/x = 1/1+x = 1

So x=0 is derivable.

i.e., f(x) is continuous and derivable at x=0.

Derivative, when the values of the two equations are equal when x=1 proves continuous; x=1 brings in two derivatives and the value of the equation is not 0

This tease is very simple, so that a=0, b=1 is a mountain disturbance.

Black circle + red circle:

The function is continuous, and the value of the middle segment of the function and the value of the left and right segments are continuous.

That is, when x = -1, x + ax-1 = -2;When x+1, x+ax-1=2. Substituting yields a=2. (There is no need for a limit, because +-1 can be taken).

Red circle: Define the fields as x>0, and x<0

y'=e^(1/x)+(x+6)e^(1/x)*(1/x²)=e^(1/x)[1-(x+6)/x²]

by y'=0, resulting in 1-(x+6) x =0

Get x -x-6 = 0

x-3)(x+2)=0

x=3, -2

When x>3 or x<-2, y'>0, the function increases monotonically;

When 0< x<3, or -2

f'(x)=4x-1/x=(4x²-1)/x=(2x-1)(2x+1)/x

Because x 0, so f'(x) is less than 0 on (0,1 2), and the function decreases monotonically.

f'(x) At (1 2, on greater than 0, the function increases monotonically.

(1) f(x) = x 2+1 x, define the domain x ≠ 0,f'(x) = 2x - 1 x 2 = (2x 3 - 1) x 2, station x = 1 2 (1 3).

f''(x) = 2 + 2/(x^3) ,f''([1 2 (1 3)] 0, minimum f[1 2 (1 3)] = 3 2 (2 3).)

2) f(x) = x + arctanx

f'(x) = 1 + 1/(1+x^2) = (2+x^2)/(1+x^2) >0

The function increases monotonically, the minimum value f(0) = 0; The maximum value f(1) = 1+ 4

You're right, since d1 and d2 are symmetrical with respect to y=x, the values of the two integrals should be equal.

In the penultimate 2 rows, the integrand of the previous integral, the exponent x has , is pulled out.

0，1）e^x²dx∫（0，x）dy

（0，1）e^x².xdx

1/2）∫（0，1）e^x².dx²

1/2）e^t|(0,1)

1/2)(e-1)

Result = e-1