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What is the equation of the tangent to the line #sf(y=ln(x^2-8)# at the point (3,0) ?

Answer 1

#y=6x-18#

The equation of the tangent is #y=6x#- 18.

To find the equation of the tangent we must find the first derivative of the function. To do this we must use the chain rule (function of a function).

This states that:

#(dy)/dx=(dy)/(dt)*(dt)/(dx)#

The outer layer of the chain is the #ln# bit. The inner layer is the #(x^2-8)# bit.

Let #t=(x^2-8)#

So #y=lnt#

So #(dy)/(dt)=(1)/(x^2-8)#

(This is because the derivative of #lnx = 1/x#)

And #(dt)/(dx)=2x#

So the product of the 2 derivatives #(dy)/(dx)# becomes--> #(dy)/(dx)=(1)/(x^2-8)*2x#

This gives the gradient of the line at a particular value of #x#. At (3,0) #x=3# so the gradient is :

#(2*3)/(3^3-8)=6#

The general equation for a straight line is #y=mx+c# where #m# is the gradient and #c# is the intercept.

To get #c# we can plug in the values we now know:

#0=6*3+c#

From which #c# = -18

So the equation of the tangent is:

#y=6x-18#

The situation looks like this:

graph{(ln(x^2-8)-y)(6x-18-y)=0 [-10, 10, -5, 5]}

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Answer 2

Charles Anctil

The equation of the tangent to the line sf(y=ln(x^2-8) at the point (3,0) is y = -3x + 9.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.