What is the equation of the tangent line of f(x)=(2x^3 - 1) / (x^2-2x) at x=3?

1 Answer
Jul 1, 2017

y=-158/9*x+211/3

Explanation:

Let

u=2x^3-1

v=x^2-2x

Then

f(x)=u/v

Using the quotient rule, which can be found in all good google searches:

f'(x)=(u'v-uv')/v^2

u'=6x

v'=2x-2

f'(x)=([6x(x^2-2x)]-[(2x^3-1)(2x-2)])/(x^2-2x)^2

Soz, I cannot be bothered simplifying this...

f'(x) is the gradient function, substituting in any x value will give us the gradient of the tangent at that point.

So let's sub in x=3:

f'(3)=([6*3(3^2-2*3)]-[(2*3^3-1)(2*3-2)])/(3^2-2*3)^2

rArrf'(3)=([18(9-6)]-[(54-1)(6-2)])/(9-6)^2=(18*3-53*4)/9=-158/9

Now we want the equation of the tangent, which is a straight line of the form:

y=mx+c

We know m=-158/9 but we need to solve for c. We need a point on the line to solve it, so sub in x=3 into the original equation to find the y value:

f(3)=(2*3^3-1)/(3^2-2*3)=(54-1)/(9-6)=53/3

So the coordinates of our point on the line are (3,53/3). Sub this point into the straight line equation to solve for c:

y=-158/9*x+c

rArr53/3=-158/9*3+c

rArr53/3=-158/3+c

rArrc=53/3+158/3=211/3

So the equation of the tangent is:

y=-158/9*x+211/3