It is proved that the limit LIM (x + y) / (X-Y) does not exist when x approaches 0 and Y approaches 0

It is proved that the limit LIM (x + y) / (X-Y) does not exist when x approaches 0 and Y approaches 0

It tends to (0,0) as a straight line y = KX (K ≠ 1)
be
lim(x+y)/(x-y)=lim(x+kx)/(x-kx)=lim(1+k)/(1-k)
The value of the limit will change with the change of K
Therefore, the limit LIM (x + y) / (X-Y) does not exist when x approaches 0 and Y approaches 0

What is the domain of the function f (x) = LG (x-x Square)?

Domain:
x-x^2>0
x^2-x

The domain of the function f (x) = LG (x squared - 3x-4) is

x ²- 3x-4＞0
（x-4)(x+1)＞0
X ＜ - 1 or X ＞ 4
Definition field (- ∞, - 1) ∪ (4, + ∞)

Definition: if f (x) has f (x0) = x0 for a number x0 in its definition field, x0 is said to be a fixed point of F (x), It is known that f (x) = AX2 + (B + 1) x + B-1 (a is not equal to 0) When a = 1 and B = - 2, find the fixed point of function f (x) If f (x) has two fixed points for any real number B, find the value range of A

(1) F (x) = x ^ 2-x-3 f (x0) = x0 x0 ^ 2-2x0-3 = 0 (x0-3) (x0 + 1) = 0 x0 = 3 or x0 = - 1
(2) Ax ^ 2 + (B + 1) x + B-1 = x ax ^ 2 + BX + B-1 = 0 △ = B ^ 2-4a (B-1) > 0, that is, B ^ 2-4ab + 4A > 0. Any value taken by B is true. G (b) = B ^ 2-4ab + 4a, the opening is upward, and there is no intersection with the horizontal axis △ = 16A ^ 2-16a

How to find the minimum value in MATLAB, such as finding the minimum value of function y = E. ^ x + 4x in the interval from - 1 to 2

x=-1:0.001:2;
y=exp(x)+4*x;
min(y)

Using MATLAB to find the maximum value of F (x) = x ^ 2-3x + 2 function in the interval [- 10,10] 1. Find the maximum value of function f (x) = x ^ 2-3x + 2 in interval [- 10,10]? 2. Find the maximum value of function f (x) = (2x ^ 2-3x + 4) / (x ^ 2 + 2x + 2) in interval [- 1,3]? 3. A factory wants to make an uncovered cuboid container with a volume of 100 cubic meters. Q: how to make the most economical materials?

syms x
f=x^2-3*x+2;
df=diff(f,x);
x0=eval(solve(df));
x=[-10 x0 10];
y=x.^2-3*x+2;
fmin=min(y)
syms x
f=(2*x^2-3*x+4)/(x^2+2*x+2);
df=diff(f,x);
x0=eval(solve(df));
x=[-1 x0(1) 3];
y=(2*x.^2-3*x+4)./(x.^2+2*x+2);
fmin=min(y)
These are the first two