Eager for advice! Regarding mathematics, the question is as follows: I hope to talk about the proces

Finding the derivative of f(x) shows that when x=0, it is meaningless at f(x) and therefore not derivative.

However, when x=0, f(x) has a value, so f(x) is continuous at x=0.

Furthermore, the answer should be D

1.Finding the defined domain can be found by the inverse function:

Let y=f(x), then y=x (1 3)+1 x=(y-1) 3 when y:(-infinity, +infinity), x:(-infinity, +infinity).

2.Let m=(y-1), then x=m 3, continuous, so x=(y-1) 3 is continuous, and the original function is also continuous.

1 [3x (-2 3)] is meaningless at x=0, therefore, not derivable.

f'(x)=(1 3)*x (-2 3)=1 (3*square of x under the cubic root sign) when x=0, the denominator is 0, which is meaningless.

It is easy to know that the domain of definition of f(x) is the entire range of real numbers.

f(-0)=f(+0)=1, so f(x) is continuous at x=0.

Deriving f(x) gives the derivative function as 1 3x (-2 3), obviously the denominator is 3x 2 and then the third power, and as the denominator, it itself cannot be zero, that is to say, the derivative function is meaningless at x=0, so the original function is not derivative at x=0.

(1) Solution: Connect to DP

cp=3bp=bc—cp=12 —3=9

ad=9 ∴ad=dp

AD DP quadrilateral ABPD is rectangular.

dp bppe dp point e coincides with point b.

2) Pass the point D as DF BC, the vertical foot is F, AD=BF=9 AB=DF=6

When the point p is on bf:

bpe +∠epd+∠dpf=180° pe⊥dp∴∠bpe +∠dpf=90°

df⊥bc∠pdf+∠dpf=90°

pdf =∠epb

△peb∽△dpf

be/pf=bp/df

cp=x be=y

bp=12—x pf=pc—cf=x—3∴y/(x-3)=(12-x)/6

y=-1/6(x2-15x+36)

When the point p is on cf, the same can be obtained: y=1 6(x2-15x+36).

Snow Purple Love Ice Rain Hello.

Let the distance be 3a, then the time is (a 30 + a 15 + a 100) and the average speed is 3a (a 30 + a 15 + a 100) = 300 11 (km).

Solution: The reason for the rise of the water surface is that an iron block is put in, so the volume volume of the rising part of the water surface is the volume of the iron block. The equation is found, and the problem is solved. The bottom area of the container is removed by the volume of the iron block, which is the height of the water surface.

The volume of the iron block is 30*10, and the bottom area of the volume is removed to 15*10;Column formula 30*10 (15*10)=2

This problem is not difficult to solve, that is, the conditions are not sufficient, and the bottom area of the iron block is not known: if the height is x cm, then (15 10-30) x = 30 (4 + x) 120x = 120 + 30x

90x=120

x = 4 3 cm.

The invariable is a cuboid container 15 cm long, 10 cm wide and 8 cm high containing 4 cm high water! Solve it accordingly!

Use a circle with a radius of one meter as the center of the circle, and surround the rest of the lamps with a radius of 1 meter, so that the "radius" of the lighting area can basically reach two meters

This question combines the practical problem with the radius and area of the circle, and tests the students' estimation ability The effect of the combination of multiple lamps in this question may not be the same as the effect of the original lamp, as long as it is basically the same

Let's start with the answer: seven.

Distributing 6 small circles evenly around the circumference of the circle, and then inserting 1 small circle in the center of the great circle, can cover the original large circle, which is the least case.

At least three to five such small lights are needed, using the overlapping method.

According to the principle of adding exponents to the same power, the original formula = 2 x * 2 3 = 3 * 8 = 24

Hope it helps.

Analysis: The principle of multiplying the power of the same base, the base is unchanged, and the exponent is added.

Solution: 2 x = 3, 2 (x+3) = (2 x) 2 =3 8=24, so the answer is 24

Comments: This question tests the knowledge of the power of the same base and belongs to the basic question.

If you have any questions, you can ask them.

Knowing 2 x = 3, find 2 (x+3) = ?

Solution: 2 (x+3) = (2 x) * 2 = 8 * (2 x ) = 8 3 = 24

Multiply the power of the same base, the base does not change, and the exponents are added; e.g. a a = a = a ; In turn, 2 (x+3) = (2 x) * (2).