Given that the slope of a tangent line of the curve y = x2-3lnx is 5, the abscissa of the tangent point is 0

Given that the slope of a tangent line of the curve y = x2-3lnx is 5, the abscissa of the tangent point is 0

The abscissa of solution tangent point is x0
Then f ′ (x0) = 5
Then y = x2-3lnx
Y ′ = [x2-3lnx] ′
=2x-3*1/x
That is, f ′ (x0) = 2x0-3 / x0 = 5
That is, 2x0 & # 178; - 5x0-3 = 0
That is, (2x0 + 1) (x0-3) = 0
The solution is x0 = - 1 / 2 or x0 = 3
That is, the abscissa of the tangent point is x0 = - 1 / 2 or x0 = 3

If the slope of a tangent line of y = x ^ 2 / 4-3lnx is 1 / 2, then the abscissa of the tangent point is

Derivation
y'=x/2-3/x
The derivative is the tangent slope
So x / 2-3 / x = 1 / 2
x²-x-6=0
x=3,x=-2
The domain x > 0, so x = 3
Substituting function to find y
So tangent point (3,9 / 4-3ln3)

How to go to the absolute value of inequality, what principles to follow and why?

It can be assumed that the discussion is divided into two cases
(1) If the formula in the absolute value is greater than or equal to zero, it will be equal to the original formula after the absolute value is removed,
(2) The formula in the absolute value is less than zero. After removing the absolute value, add a negative sign before the original formula
In the calculation, we usually multiply a square number, which can avoid the change of inequality symbol