Solving mathematical continuous function problems must process .

As can be seen from the title, when x<0 or x>0, the function is continuous, so we only need to discuss the continuity of the function when x=0.

Since the expressions are different on both sides of 0, the left and right limits of 0 are considered.

lim【x→0+】f(x)=lim【x→0+】(3x²-2x+k)=k

lim [x 0-] f(x) = lim [x 0-] sin2x x=2 so that the function is continuous at x=0 point, it must be required.

lim【x→0+】f(x)=lim【x→0-】f(x)=lim【x→0】f(x)=f(0)=k

Thus we can see that when k=2, the function is continuous at x=0, and thus continuous on the defined domain!

If you don't understand, you can ask, if it helps, please choose satisfied!

Hello! This should be the answer (you can't use a formulator here, I'll give you a general answer):

1) When x tends to 0 on the left, lim f(x)=lim (sin 2x) x =lim [(sin 2x) 2x]*2 its value tends to 2, that is, the value is 2;

2) When x tends to 0 on the right, f(x)=3x-2x+k, then x can take a value of 0, i.e., f(0)=k, 3) if f(x) is to be continuous in its defined domain, i.e., (1) and (2) are equal, f(0)=k=2If it is not equal, the function is a breakpoint function that is contiguous and discontinuous within its defined domain.

x->0-, the left limit is 2

x->0+, the right limit is k

To make the function continuous at x=0, then k=2

The answer to the question was a typographical error and proved to be inadequate.

The proof is given below.

f(x) is continuous in [0,2a], and f(a+x) is obtained by the composite of two continuous functions y=f(u) and u=a+x, so continuous, the continuous interval of f(u) is [0,2a], so the continuous interval of f(a+x) is [-a,a], so the continuous interval of f(x)=f(a+x)-f(x) should be [0,a].

f(a)=f(2a)-f(a), f(0)=f(a)-f(0)=f(a)-f(2a)=-f(a)

If f(a)=0, then f(2a)=f(a),a[0,a] conclusion is true.

If f(a) ≠0, then f(a)f(0)<0, by the zero theorem of continuous functions, exists (0,a), such that f( )0, i.e.

f( )f( +a) conclusion is valid.

The continuous interval of the function is the domain of the definition of high numbers, doesn't the title say that the continuous interval of f(x) is [0,2a]?

Suppose max=m

min=m(t1+t2+...)tn)m=i.e., m<=f(c)<=m

The value of f(c) is between the maximum and minimum values, which is known by the intermediate value theorem.

There is at least a little bit of c on [a,b] such that f(c) = tif(x1)+....tnf(xn)

Since both functions are elementary functions, they must be continuous in their respective domains, and now it is necessary to ensure that they are continuous at the demarcation point. Substituting x=0 into two functions, the two should be equal, so the following equation is obtained: 0+1=a+2*0, so the solution is a=1

4, let t be an infinitesimal quantity, then f(x)=x 3, for any x there is f(x+t)=(x+t) 3=x 3+3x 2t+3xt 2+t 3=f(x)+3x 2t+3xt 2+t 3=f(x)+m

When t tends to 0, and m tends to 0 from the above equation, then the limit f(x+t)=f(x), then f(x) is continuous.

5. In the same 4 questions, it is easy to obtain f(x) which is continuous at x= 1, and it is easy to obtain f(1)=2, and the left limit = right limit = 2=f(1), then f(x) is continuous in the field of real numbers.

At x=1, the left limit (equal to 0) and the right limit (=2 1+1=3) are not equal, so they are discontinuous at x=1;

At x=2, the left limit (=2 2+1=5) and the right limit (=f(2)=1+2 2=5), so at x=2 it is continuous;

No, because when x=1, f(x)=0 and 3 when x=. f(x)=5

x 1, and x 2, the first derivatives of the two segments before and after the function are not equal and therefore not continuous.

At x=1, the left limit (equal to 0) and the right limit (3) are not equal and discontinuous;

At x=2, the left limit = the right limit = f(2) = 5, continuous;

I'll go ahead and write gmlonger's answer in detail:

The continuous function must be continuous at all points, and when x<3 and x>3, the function is continuous, because they are quadratic and primary, respectively, and there is no problem. The problem is that continuity at x=3 cannot be guaranteed. To ensure continuity at this point, we must first ensure that the function has an limit at this point, and the sufficient and necessary conditions for the function to have a limit are that the left and right limits of the function exist and are equal.

Left limit: f(x)(when x approaches 3 from left) = lim (x--3-) x 2-1) = 8;

Right limit: f(x)(when x tends to 3 from the right) = lim (x--3+) 2ax) =6a;

The two limits must be equal, so there is a = 4 3

Secondly, it is not enough to have a limit, it is also necessary to ensure that the limit value at this point is equal to the value of the function, because when a = 4 3, f(3) = 2a*3 = 8, it is indeed equal to the limit above, so the answer is a = 4 3