The line y = 2x + 2 and X, Y axis intersect with a, B respectively to find the AOB area of triangle

The line y = 2x + 2 and X, Y axis intersect with a, B respectively to find the AOB area of triangle

analysis
When x = 0
y=2
y=0
x=-1
So area
=1/2xy
=1/2x2
=1

Finding the monotonicity of definition range of function y = (2x + 1) / (3x-4)

y=(2/3)[(x+1/2)/(x-4/3)]
=(2/3)[1+(11/6)/(x-4/3)]
=2/3+(11/9)/(x-4/3),
Its definition field is x ≠ 4 / 3, and its value field is y ≠ 2 / 3,
It is monotonic decreasing at X4 / 3

Cut the cuboid which is 60cm long into three sections, so that the surface area is increased by 180 square centimeters. What is the original volume of the cuboid

180 / 4 * 60 = 2700 cm3

If the equation x ^ 2 + MX + 1 = 0 about X has two unequal real roots, then the value range of real number m is

m^2-4＞0
M ＜ - 2 or m ＞ 2

If a and B are reciprocal, C and D are opposite, M is the absolute value of - 3
Finding the value of 2ab-3 (c + D) △ 5 + M

That is ab = 1
c+d=0
m=|-3|=3
So the original formula is 2 * 1-3 * 0 △ 5 + 3
=2-0+3
=5

Given that f (x) = x / (AX + b), (a, B are constants, and ab ≠ 0), and f (2) = 1, f (x) = x has a unique solution, find the analytic expression of F (x)

Given that the center of circle x square + y square - 4x-2y-6 = 0 is on the straight line ax + 2by-2ab = 0, a > 0, b > 0, find the minimum value of ab
It's very easy to know this problem. First, find out the coordinates of the center of the circle, then substitute them into the linear equation, and then? How to use the mean value theorem?

X square + y square - 4x-2y-6 = 0
(X-2) ^ 2 + (Y-1) ^ 2 = 11, the center of the circle is (2,1) on the straight line ax + 2by-2ab = 0, then
2a+2b-2ab=0
a+b-ab=0
a+b=ab≥2√ab
√ab≥2
ab≥4
If and only if a = b = 2, take "="
The minimum value of AB is 4

Given the function f (x) = 2-x2, G (x) = X. if f (x) * g (x) = min {f (x), G (x)}, then the maximum value of F (x) * g (x) is______ (Note: Min is the minimum value)

From the meaning of the question, we can draw a function image that meets the conditions, such as f (x) * g (x) = 2 − x2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; - 2 − x2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X ≥ 1 x < 1

Taylor formula for higher derivative
F (x) = x ^ 3 · SiNx & nbsp; & nbsp; & nbsp; use Taylor formula to find the sixth derivative when x is equal to 0.

Using the Taylor expansion of SiNx, SiNx = x-x ^ 3 / 3! + x ^ 5 / 5! - x ^ 7 / 7! +..., so
f(x)=x^4-x^6/3!+x^8/5!-x^10/7!+...
So we know that f ^ (6) (0) / 6! = - 1 / 3
f^(6)(0)=-6!/3!=-120.

Let f (x) defined on R satisfy f (x) · f (x + 2) = 13, if f (1) = 2, then f (99) = ()
A. 13B. 2C. 132D. 213

∵ f (x) · f (x + 2) = 13 and f (1) = 2 ∵ f (3) = 13F (1) = 132, f (5) = 13F (3) = 2, f (7) = 13F (5) = 132, f (9) = 13F (7) = 2, ∵ f (2n − 1) = 2 & nbsp; n is odd 132 & nbsp; n is even, f (99) = f (2 × 100 − 1) = 132, so C is selected