b) Find the gradient of y= 2ln x - x^2 when x = 2?

Question

and b) Find the coordinates of the turning point

Answer 1

Part a)
f ( x ) = 2 ln x - x ²
f `( x ) = 2 / x - 2 x
f `( 2 ) = 1 - 4
f `( 2 ) = - 3
m = - 3

Part b)
f ( 2 ) = 2 ln 2 - 4
f ( 2 ) = - 2.61
Turning point (2 , - 2.61)

Answer 2

The gradient is the ∂y/∂x:
so ∂y/∂x = 2/x - 2x
Now when x = 2
∂y/∂x(2) = (2/2) -2(2) = 1 - 4 = -3