The function f (x) = X-1 / 2aX ^ 2-ln (1 + x), where a ∈ R (Ⅰ) if x = 2 is the extreme point of F (x), find the value of A

The function f (x) = X-1 / 2aX ^ 2-ln (1 + x), where a ∈ R (Ⅰ) if x = 2 is the extreme point of F (x), find the value of A

f'(x)=1-ax-1/(1+x)
From the meaning of the question, x = 2 is the extreme point, that is, f '(2) = 1-2a-1 / 3 = 0
Solution: a = 1 / 3

Given the function f (x) = ln (1 / 2 + 1 / 2aX) + x ^ 2-ax (a is a constant, a > 0) (1) when a = 1, find the tangent equation of function f (x) at x = 1 (2) When y = f (x) obtains the extreme value at x = 1 / 2, if the equation f (x) - B = 0 about X has exactly two unequal real roots on [0,2], find the value range of real number B (3) For any a ∈ (1,2), there is x ∈ [1 / 2,1], so that the inequality f (x) > m (a ^ 2 + 2a-3) holds, and the value range of the real number m is obtained

0

The function f (x) = ln (1 / 2 + 1 / 2aX) + x ^ 2-ax. (a is a constant, a > 0) (1) if x = 1 / 2 is an extreme point of function f (x), find the value of A (2) Verification: when 0

1) If x = 1 / 2 is an extreme point of function f (x), find the value of A
f'(x)=1/[1/2+1/(2ax)]+2x-a
f'(1/2)=0=1/(1/2+1/a)+1-a=2a/(a+2)-a+1=(-a^2+a+2)/a=(a+1)(2-a)/(2a)
a=2 or a=-1

The known function f (x) = ln (1 / 2 + 1 / 2aX) + x ^ 2-ax. (a is a constant, a > 0) is verified: The known function f (x) = ln (1 / 2 + 1 / 2aX) + x ^ 2-ax. (a is a constant, a > 0) is verified when 0

Ax is the denominator? Molecules?

The derivative of Ln (2-x). A little confused According to the formula (LNX) '= 1 / x, shouldn't the derivative of Ln (2-x) be 1 / 2-x Why is the answer 1 / X-2

In fact, it takes two steps to find the derivative, (LNX) '= (1 / x) * (x)' = 1 / x, so (LN (2-x)) '= (1 / 2-x) * (2-x)' = 1 / X-2

If there are straight lines and curves passing through points (1, 0), y = X3 and y = AX2 + 15 4x-9 are tangent, then a is equal to _

From y = x3 ⇒ y '= 3x2, let the tangent equation at any point (x0, X03) on curve y = X3 be y-x03 = 3x02 (x-x0), (1, 0) be substituted into the equation to obtain x0 = 0 or x0 = 3
two
① When x0 = 0, the tangent equation is y = 0, then AX2 + 15
4x-9=0，△=(15
4)2-4a × (-9)=0⇒a=-25
sixty-four
② When x0 = 3
2, the tangent equation is y = 27
4x-27
4. By
y=ax2+15
4x-9
y=27
4x-27
four ⇒ax2-3x-9
4=0，△=32-4a(-9
4)=0⇒a=-1∴a=-25
64 or a = - 1
So the answer is: - 25
64 or - 1