Simple function questions, very simple function questions

Derivative of f(x).

f'(x)=3x^2+2bx+c

Since f(x) is a continuous function, and the problem shows that f(x) has a turn at x=0, so f'(0)=0

Get c=0

Since f(x) is continuous and is a subtractive function on (0,2), so f'(2) <=0, i.e., 12+4b<=0 to get b<= -3

Bring x=2 into f(x)=0 to get.

8+4b+0+d=0

This gives d=-8-4b

f(1)=1+b+c+d = 1+b+0+(-8-4b) = -7-3b

Because b<= -3

So -7-3b>=2

So f(1)>=2

f(2)=0,∴d=-4(b+2), f'The two roots of (x)=3x 3+2bx=0 are x1=0 and x2= -2b 3

The function f(x) is a subtraction function on [0,2], x2 = -2b 3 2

b≤-3f(1)=b+d+1=b-4(b+2)+1=-7-3b≥2

f(2)=2

Then 8+4b+2c+d=0 .1

Derivative of f(x) yields f(x) derivative = 3x 2 + 2bx + c and f(0) derivative = 0

then c = 0 is substituted into 1.

8+4b+d=0

i.e. d = -8-4b

is a subtraction function on (0,2).

3x^2+2bx<0

then 3x+2b<0

When x=2, 6+2b<=0

3+b<=0

f(1)=1+b+d=-7-3b=2-3(3+b)>=2

Solution: Let t = 3 x, then t 0, then f(x) = t -k + 1) t + 2 to be positive when x belongs to r.

Then t -k + 1)t + 2 0 is constant, i.e.: t +2 (k +1)t

t² +2)/t ＞ k + 1

Hence k t + 2 t - 1

For t + 2 t, when t = 2, t + 2 t has a minimum value of 2 2 and when k is less than the minimum value of t + 2 t - 1, the original formula is constant, k 2 2 - 1

Let t=3 x be functionalized as g(t)=t 2-(k+1)t+2 (t>0).

The function f(x)=3 2x-(k+1)*3 x+2, when x r, f(x) is everstable at zero.

i.e., g(t)=t 2-(k+1)t+2>0 whose discriminant formula is less than zero, i.e.

k+1)^2-4*2<0

k^2+2k-7<0

2 roots 2-1

There are three conditions for the Law of Lobida:

1. The function of the numerator and denominator approaches 0 or infinity at the same time;

2. It is derivable in the decentric neighborhood of the point, and the derivative function of the denominator function is not 03, and then take the limit after the derivative, and then take the limit, and find a constant a, then the original limit value is a Obviously, not every 0 to 0 limit can use the Lobida rule, and your process should be to verify that the conditions for using Lobida are not met, and it is obvious that the condition 3 is not met. So the conclusion is that this limit cannot be found using Lopida's law. The Law of Lobida is not a panacea.

No, because x 2 x is infinitesimal and sin(1 x) is a bounded function. Then multiply the infinitesimal quantity according to the bounded function as an infinitesimal quantity, so the final limit is 0

y=(1+to the power of x.

That is, y= to the power of x.

Let it be equal to, x = the logarithm of the true number with the base.

Let y1=m(x+1) and y2=n x

Sharp crack such as the source of y=y1+y2=m(x+1)+n x put x=1, y=0, when x=4, y=9

Bring in the silver enlightenment.

2m+n=0,5m+n/4=9

The solution yields m=2 and n=-4

y=m(x+1)+n/x=2x+2-4/x

For this, use trigonometric functions to find the values of a and b, and then substitute them into algebraic formulas, just like the first floor.

f(x)=2x+1

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

Since f(x) defines the domain as [1,3], x-1 [1,3] in f(x-1) gives x [2,4].

So the answer is B

is a multiple-choice question about functions.

The test is about function expressions and definition fields, and it depends on whether you understand the concept of functions f(x)=2x+1

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

Since f(x) defines the domain as [1,3], x-1 [1,3] in f(x-1) gives x [2,4].

So the answer is B