question number 5

Question

For the function f(x) = 5x^2 − 4x:

a. find f^inverse (x)

b. find f^inverse (x) by evaluating the limit of the difference quotient for f(x).

*i got 5h+10x-4 as the limit as h approaches 0, but apparently i didn't finish it all the way... not sure where i went wrong

c. find the equation of the tangent line to f(x) at x = 1.

Answer 1

y= 5x^2 - 4x

a) to find the inverse, f^-1(x)

switch x and y, then solve for the new y

x = 5y^2 -4y

5y^2 -4y -x = 0

use the quadratic formula

y = 4/10 + or - (1/10)sqr(16 - 4(5)(-x))

y = [4 + or - sqr(16+20x)]/10

that's the inverse but it's not an inverse function, since there's 2 values of y for one value of x

It's an inverse relation

c) tangent line:

y=5x^2 - 4x has slope = the derivative

y' = 10x -4, at x=1

y'(x) = y'(1) = 10-4 = 6 = slope of the tangent line at x=1

at x=1 y = 5x^2 -4x = 5-4 = 1. It's the point (1,1)

y=mx +b = 6x + b, plug in the point (1,1) to solve for b

1 =6 + b

b =1-6 =-5

the tangent line on the curve at x=1 is

y = 6x -5

b) difference quotient for y= 5x^2 -4x

f(x+h) - f(x) all over (x+h)-x = [f(x+h)-f(x)]/h

= [5(x+h)^2 -4(x+h) - (5x^2 -4x)]/h

= (5x^2 +10xh + 5h^2 -4x -4h -5x^2 +4x)/h the x^2 and x terms cancel

= (10xh +5h^2-4h)/h = 10x +5h -4

let h go to zero,

= 10x +5(0) -4 = 10x -4 = the derivative of 5x^2 -4x