Find (f^-1)'(2) if f(x)=2x^2-3x for x>0

Answer 1

Exchange the roles of x and y: x = 2y²-3y

Use quadratic formula to solve for y: y = (3+sqrt(9+8x))/4

dy/dx = 4/sqrt(9+8x) and evaluated at x = 2 this is 4/5

Answer 2

3 things need to be done.....

1) inverts the function

2) takes the derivative

3) plugs in x=2

1) finds the inverse

y = 2x^2 - 3x

x = 2y^2 - 3y <--- swaps X and Y

0 = 2y^2 - 3y - x <--- writes as quadratic

quadratic formula says, with A=2, B=-3 and C = -x

y = [3 +or- sqrt ( 9 - 4(2)(-x))] / 4

= [ 3 +or- sqrt( 9 + 8x)]/4

= 1/4[ 3 +or- sqrt( 8x+9) ]

= 1/4 [ 3 +or- (8x+9)^(1/2) ]

Part 2) finds the derivative

1/4 [ +or- (1/2) ( 8x+9)^(-1/2) ] ( 8)

three differentiates to zero, power rule causes 1/2 to go in front, making

the exponent -1/2; then we must use the chain rule, which says the derivative

of the linear function on the inside is 8

the fractions cancel the 8...

the derivative is +or- (8x+9)^(-1/2)

part 3) x=2

+or- (8*2 + 9)^(-1/2) =

+or- (25)^(-1/2) =

+or- 1/ sqrt(25) =

+or- (1/5)