The maximum value of the function y = lnxx is () A. e-1B. eC. e2D. 103

The maximum value of the function y = lnxx is () A. e-1B. eC. e2D. 103

Let y ′ = (LNX) ′ x − LNX · x ′ x2 = 1 − lnxx2 = 0, x = e, when x ＞ e, y ′＜ 0; when x ＜ e, y ′＞ 0, y maximum = f (E) = 1E, there is only one extremum in the domain, so ymax = 1E, so the answer is a

Find the maximum value of the function y = LNX / X

If y '= (1-lnx) / (x * x) is derived from the function so that y' = 0 is x = e, we should also consider the point x = 0 where the derivative does not exist. The domain of definition is x > O, and x = 0 is not in the domain of definition when 0

Given the function f (x) = LNX of X, find the maximum value of FX

The function definition field is (0, + ∞)
f'(x)=(1-lnx)/x²
Let f '(x) > 0, then 0

Finding the maximum value of a given function f (x) = LNX / x-x

f'(x)=1/x²-lnx/x²-1,
When x = 1, f '(x) = 0,
When 0 ＜ x ＜ 1, 1 / X & # 178; - 1 ＞ 0, - LNX / X & # 178; > 0,
f'(x)=(1/x²-1)-lnx/x²＞0,
When x ＞ 1, f '(x) ＜ 0,
So f (x) increases first and then decreases on (0, + ∞),
When x = 1, the maximum value is - 1

Calculate 111 12004 1-222 21002 2 = a × a, find a

Because 11-2 = 3 × 31111-22 = 33 × 33111111-222 = 333 × 333 12004 1-222 21002 2 =, so a = 333 3 (1002 3)

The distance between the center of a circle and the chord is 1cm. The chord length is ()

Yes, 4 √ 2 is connected with OC Pythagorean Theorem 1 & # 178; + 5 & # 178; = 8 = 2 √ 2 in x 2

Find the limit (2x3-3x + 4) / (8 + 5x2-3x3)

Original formula = LIM (x - > infinity) [2 - (3 / x ^ 2) + (4 / x ^ 3)] / [(8 / x ^ 3) + (5 / x) - 3]
=2/(-3)
=-2/3

If the length of one side of a parallelogram is 10, the length of two diagonals can be ()
A. 4 or 8b. 6 or 8C. 8 or 10d. 10 or 12

As shown in the figure: ∵ quadrilateral ABCD is a parallelogram, | OA = OC = 12ac, OB = od = 12bd, if BC = 10, according to the triangle trilateral relationship, we can get: | ob-oc | 10 < ob + OC. A, OB = 2, OC = 4, | ob + OC = 6 < 10, can not form a triangle, so this option is wrong; B, OB = 3, OC = 4, | ob + OC = 7 < 10, can not form a triangle, so this option is wrong; C, OB = 4, OC = 5, | ob + OC = 9 < 10, not correct It can form triangle, so this option is wrong; D, OB = 5, OC = 6, х ob + OC = 11 > 10, oc-ob = 1 < 10, can form triangle, so this option is correct. So select D

1. 30% x = 420 2. X-20% x = 4.8 3.120% x + 40% x = 12.8

30%x=420
x=420÷30%
x=420×100/30
x=1400
x-20%x=4.8
80%x=4.8
x=4.8÷80%
x=6
120%x+40%x=12.8
160%x=12.8
x=12.8÷160%
x=8

Find the average change rate of (x ^ 2 + 1) from x0 to x0 + △ x under y = root sign

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