If the zeros of function f (x) = 3x-7 + LNX are in the interval (n, N + 1) (n belongs to n), then n= I want to know this in the following process: Let F1 (x) = 3x-7, F2 (x) = LNX. F (x) = F1 (x) + F2 (x) When x = 2, F1 (x) = - 10 So when f (x) = 0, X must be in the interval (2,3) N=2 Why does it have to be in the (2,3) interval? Isn't it written clearly that when = 2, it is ＜ 0? Why does it become n = 2 when = 0? This idea can't be changed

If the zeros of function f (x) = 3x-7 + LNX are in the interval (n, N + 1) (n belongs to n), then n= I want to know this in the following process: Let F1 (x) = 3x-7, F2 (x) = LNX. F (x) = F1 (x) + F2 (x) When x = 2, F1 (x) = - 10 So when f (x) = 0, X must be in the interval (2,3) N=2 Why does it have to be in the (2,3) interval? Isn't it written clearly that when = 2, it is ＜ 0? Why does it become n = 2 when = 0? This idea can't be changed

If the function is continuous, there must be a point where the value of the function is 0
Because the zero point is in the interval when n = 2
Not at 3
How do you think about it
Since the function value of 2 is less than 0
Greater than 0 at 3
Function must be an increasing function in this section
Of course, take the function value on the left
Only in this way can zero be in this interval
Do you understand
Why is it 2 instead of 3? I ask you, after that, will this kind of topic take smaller value? No big one? Please say again, I didn't see hi! I revised my question later. Please show it to me
You can refresh one... And unfold it
Because the zero point is in the interval when n = 2
Not at 3
How do you think about it
Since the function value of 2 is less than 0
Greater than 0 at 3
Function must be an increasing function in this section
Of course, take the function value on the left
Only in this way can zero be in this interval
Do you understand
Why 2 is not 3? Ask: I ask you, after that this kind of question all take small? No big one?
The greatest divisor of a number is 30, the least common multiple of the number is (), and the prime factor of decomposing the number is ()
The greatest divisor of a number is 30, and the least common multiple of the number is (30). The prime factor of decomposing the number is (30 = 2 × 3 × 5)
The least common multiple of this number is 30, and the prime factor is 2x3x5
If the zeros of the function f (x) = 3x-7 + ln & nbsp; X lie in the interval (n, N + 1) (n ∈ n), then n=______ .
Because f (1) = - 4 < 0, f (2) = ln & nbsp; 2-1 < 0, f (3) = 2 + ln & nbsp; 3 > 0, and f (x) is an increasing function on (0, + ∞), so in the interval (2,3), n = 2, so the answer is: 2
There are () divisors of 36. Decompose 36 into prime factor ()
The greatest common divisor of 72 and 90 is () and the least common multiple is ()
The largest internal angle in a triangle may be 59 degrees
Divisor of 36 = 1,2,3,6,12,18,36
36 decomposition quality factor = 2 * 2 * 3 * 3
The greatest common divisor of 72,90 = 18
Least common multiple = 360
If the maximum internal angle is 59 degrees, then the sum of the maximum internal angles of the triangle = 177 degrees < 180 degrees
So wrong!
If the derivative of function f (x) at x = a is a, find the value of limh tending to 0f (a + 4h) - f (a + 5H) / h
f'(x) = lim(h->0) [f(x+h) -f(x) ] /hf'(a) =lim(h->0) [f(a+h) -f(a) ] /hlim(h->0) [f(a+4h) -f(a+5h)] /h=lim(h->0) [ (f(a+4h) -f(a)) - ( f(a+5h)-f(a)) ] /h= lim(h->0) [f(a+4h)-f(a)] /h - lim(h->0) [f(a+...
-A
The sum of three continuous natural numbers is 18. The greatest common divisor of these three natural numbers is______ The least common multiple is______ .
According to the meaning of the question, the average of the three natural numbers is: 18 △ 3 = 6, then the middle of the three continuous natural numbers is 6, 6-1 = 5, 6 + 1 = 7, so the three continuous natural numbers are: 5, 6, 7; 5, 6, 7 are mutually prime, so their greatest common factor is 1, and the least common multiple is: 5 × 6 × 7 = 210
Let f (x) have two continuous derivatives, and (x - > 0) Lim [f (x) - A] / [e ^ x ^ 2-1] = 0, (x - > 0) Lim [f '(x) + 1] / [1-cosx] = 2, then
Why can't we use lobida twice for the first condition? Isn't it still type 0 / 0 after the first lobida?
When X - > 0, 0.5 * x ^ 2 is infinitesimal. If the limit of LIM [f '' (x) + 1] / 0.5 * x ^ 2 exists and is equal to 2, then f '' (x) + 1 must also be infinitesimal, that is, Lim [f '' (x) + 1] = 0
If the natural number C is 5 times of B, then the least common multiple of B and C is______ The greatest common divisor is______ .
If the natural number C is 5 times of B, then the least common multiple of B and C is C and the greatest common factor is B; so the answer is: C, B
If f (x) has continuous second derivative, f (0) = 0, f '(0) = 1, f' '(0) = - 2, then (x → 0) LIM (f (x) - x) / x2 =?
Finally, divide by the square of X, that two dozen is a little big
If the following limits tend to zero, I will not repeat x → 0
The function f (x) has continuous second derivative
Both f '(x) and f' '(x) exist
You can use the law of lobita
LIM (f (x) - x) / X2 (type 0 / 0)
=LIM (f '(x) - 1) / 2x (type 0 / 0)
=limf''(x)/2
=f''(0)/2
=-1
Derivation of LIM (f (x) - x) / x ^ 2
=lim(f'(x))/2x
=limf''(x)/2
=-1
LIM (f (x) - x) / x ^ 2, because f (0) = 0, the molecular denominator of the limit is 0, so the derivation of the molecular denominator is obtained by using the robit rule. Let's get LIM (f '(x) - 1)) / (2x) and then take x = 0, which is the same case. Let's use the robit's law to seek the derivatives of the numerator and denominator respectively, and get limf' '(x) / 2 and then take x = 0, - 2 / 2 = - 1
It is known that the sum of two natural numbers is 54, and the difference between their least common multiple and greatest common divisor is 114
Let two natural numbers be x and y, x = AXB, y = CXB, and a, B, C are all positive integers, then AB + CB = 54 = B (a + C) = 2x3x9abc-b = 114 = B (ac-1) = 2x3x19. Because a, B, C are all positive integers, B may be 2 or 3 or 6
24 and 30 questions: can you be more detailed