Given the function f (x) = ax ^ 2 + blnx, when x = 1, it has extremum 1. Find the value of A.B and the monotone interval of the function

Given the function f (x) = ax ^ 2 + blnx, when x = 1, it has extremum 1. Find the value of A.B and the monotone interval of the function

f(x)=ax^2+blnx
f'(x)=2ax+b/x
x=1
f'(1)=2a+b=0
f(1)=a=1
b=-2
f'(x)=2x-2/x(x>0)
f'(x)>0
2x^2-2>0
x>1
[1, R] monotone increasing
(0,1) monotone decreasing

Find the extremum of function y = (X & # 178; - 1) &# 178; + 1
But the answer is only the minimum f (0) = 0???

Find the derivative. Let the derivative = 0, get the abscissa of the extreme point and "inflection point". Substitute y, you can get the maximum value of Y. first order function, straight line, one direction; second order function, parabola, two directions, V or inverted V shape; third order parabola, remove simple, three directions, n or inverted n shape; fourth order parabola, remove simple, four directions

Find the extremum of function y = (X & # 178; - 2) / 2 (x-1) &# 178
The answer is that the maximum is - 3 / 8, and there is no minimum. Ask for detailed explanation

Y '= 1 / 2 * [2x * (x-1) ^ 2 - 2 (x ^ 2 - 2) (x-1)] / (x-1) ^ 4 = 1 / 2 * (- 2x ^ 2 + 6x - 4) / (x-1) ^ 4 = - (x-1) (X-2) / (x-1) ^ 4 = - (X-2) / (x-1) ^ 3 when x > 2, the derivative is negative (note the negative sign before), the function is monotonous

What is the answer of 2005, 2005 × 2004-2004, 2004 × 2005? The teacher said that there must be two steps,

20052005*2004-20042004*2005=
2005*2004*10001-2005*2004*10001
=0

As shown in the figure, ⊙ P and ⊙ o intersect at two points a and B, ⊙ P passes through the center O, and point C is any point of AB on the superior arc of ⊙ P (not coincident with points a and b), connecting AB, AC, BC and OC. (1) point out an angle equal to ∠ ACO in the figure; (2) when point C is at what position on ⊙ P, is the straight line CA tangent to ⊙ o? Please explain the reason; (3) when ∠ ACB = 60 °, what is the size relationship between the radius of two circles? Please state your reasons

(1) In ⊙ o, ∵ OA = ob, ∵ OA = ob, ∵ ACO = ∠ BCO; (2) connect OP, and extend the intersection with ⊙ P at point D. if point C is at point D, the straight line CA is tangent to ⊙ O. reason: connect ad, OA, then ∵ Dao = 90 °, OA ⊥ Da ⊙ Da is tangent to ⊙ o

(- 15) divided by (1 / 3-1 / 2) multiplied by 6

(- 15) divided by (1 / 3-1 / 2) multiplied by 6
= 15 ÷ (1/2- 1/3 )
= 15 x 6
= 90
Benefactor, I see your bones are strange,
He is a man of noble bearing and wisdom,
He is a unique talent in the Wulin
If you devote yourself to study, you will become a great tool in the future,
I have a little test, please click next to the answer
"Choose as satisfactory answer"

As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connect BC and De, intersect at point F, BC and ad intersect at point g. verification: BC = De

In △ cab and △ ead, ab = ad, BAC = DAE, cab = SAS, BC = De

10 / 2 of 13 and 19-5 of 22 * 11 of 13 + 7 + 1 of 5 * 22 of 63

10 / 2 of 13 and 19-5 of 22 * 11 of 13 + 7 + 1 of 5 * 22 of 63
=13 / 10 * 63 / 22-5 / 2 * 13 / 11 + 7 + 5 / 1 * 63 / 22
=(10 / 13 + 1 / 5) * 22 / 63-22 / 65 + 7
=63 of 65 * 22 of 63-22 + 7 of 65
=22 / 65-22 / 65 + 7
=7

Finding determinant of block matrix
0 A
B 0 is not a determinant of a matrix - | a | B|

It's not necessary that a is of order k, B is of order n, and then we have to multiply it by the K + n power of minus one

How to divide 52 by 6.5 and multiply 26 by 0.37

52 △ 6.5 + 26 × 0.37 multiply and divide first, then add and subtract