Finding the minimum value of function y = 3x + 1 / (2x ^ 2) (x > 0) thank you

Finding the minimum value of function y = 3x + 1 / (2x ^ 2) (x > 0) thank you

Y = 1.5x + 1.5x + 1 / 2x ^ 2 > = 3 (1.5x * 1.5x * 1 / 2x ^ 2) cube root = 3 * (9 / 8) cube root = (3 / 2) * 9 cube root
When 1.5x = 1.5x = 1 / 2x ^ 2, take the equal sign
In this case, the cube root of x = (1 / 3) conforms to x > 0
So we can get the equal sign
So the minimum value is the cube root of (3 / 2) * 9

How to find the minimum value of function y = | X-1 | + | 2x-1 | + | 3x-1 | +... + | 10x-1 |?
This paper is from Zhenhai District of Zhejiang Province in 2011. It is very troublesome to solve this problem by using the zero point segmentation method. Is there a simple method? If the problem is further extended: find the minimum value of the function y = | X-1 | + | 2x-1 | + | 3x-1 | +... + | 2012x-1 |?

When x > = 1, y = 55x-10; when x 1 / 7, y increases with the increase of X;
When n > = 7, the slope of y = 1423, the slope of Y

Function y = x ^ 3-3x ^ 2-9x (- 2 ＜ x ＜ 2) yes? (answer max 5, no min) why?

The derivation is y '= 3x ^ 2-6x-9
Let y '= 3x ^ 2-6x-9 = 0 to find two roots
One of them is not in the range (- 2 < x < 2), so it can't be used

a> 1, b > 1, log2 (a) * log2 (b) = 64, find the minimum value of log2 (AB)

A > 1, b > 1, log2 (a) > 0, log2 (b) > 0
Log2 (a) * log (b) = 64, then log2 (AB) = log2 (a) + log (b) ≥ 2, root 64 = 16
If and only if log2 (a) = log2 (b) = 8, that is, a = b = 256, the minimum value is 16

1 / 6 + x = 7 / 36 solution

1/6+x=7/36
x=7/36-1/6
x=1/36

In n positive integers 1,2,..., N, there are p prime numbers, Q composite numbers, m odd numbers and n even numbers, then (P-M) + (q-n)=

(p-m)+(q-n)
=(p+q)-(m+n)
P + Q is all numbers except 1 (n-1 numbers)
M + n is all numbers (n numbers)
therefore
(p+q)-(m+n)=(N-1)-N=-1

1/1024 + 1/512 + 1/256 +.+ 1/2 + 1 + 2 + 4 + 8 +.+512=?

It can be divided into two equal ratio sequence, the 10th power of 2 is 1024, the 9th power of 2 is 512, the 8th power of 2 is 256, and so on. The sum formula of equal ratio sequence is Sn = A1 (1-Q ^ n) / (1-Q) = (A1 an * q) / (1-Q) (Q ≠ 1). Sn = (1 / 2-1 / 1024 * 1 / 2) / (1-1 / 2) = 1023 / 1024 or: 1 / 2 + 1 / 4 + 1 / 8 +

How much is 1 and 6 1 + 2 and 12 1 + 1 + 3 and 20 1?

After a look, it seems that there is no easy way. Just do it like this = (1 + 2 + 1 + 3) and (1 / 6 + 1 / 12 + 1 / 20) = 7 and (1 / 4 + 1 / 20) = 7 and 3 / 10

We know the quadratic power of a + 5A + 1 = 0

A + 1 / a = (quadratic power of 1 + a) / A
Then, the quadratic power of a + 5A + 1 = 0, that is, (the quadratic power of 1 + a) = - 5A is substituted into the above formula, and the result = - 5

Find the symmetry axis, vertex coordinates and the intersection coordinates of X axis of the image with function y = 4x2 + 24x + 35

∵ y = 4x2 + 24x + 35 = 4 (x2 + 6x) + 35 = 4 (x2 + 6x + 9-9) + 35 = 4 (x + 3) 2-1, ∵ the axis of symmetry is a straight line x = - 3, and the vertex coordinates are (- 3, - 1). By solving the equation 4x2 + 24x + 35 = 0, we get X1 = − 52, X2 = − 72. Therefore, the coordinates of its intersection with the X axis are (− 52, 0), (− 72, 0)