At what point on the curve y=[ln(x+4)]^2 is the tangent horizontal?

Question

Am I suppose to derive first in order to find the points?

Answer 1

y = [ ln(x + 4) ]^2

To find out where the tangent is horizontal, you want to determine where the slope is zero.  The first derivative tells you the slope at a given point.  

Differentiating, we get

dy/dx = 2 [ ln(x + 4) ] (1/(x + 4))

Which reduces to

dy/dx = 2 [ ln(x + 4) ] / (x + 4)

Make dy/dx = 0 (because we want to know at which points of x make the slope 0).

0 = 2 [ ln(x + 4) ] / (x + 4)

Divide by 2,

0 = ln(x + 4) / (x + 4)

This is equal to zero when the numerator is equal to zero.  Equating the numerator to zero gives us

ln(x + 4) = 0

And if we solve for x, we get

x + 4 = e^0

x + 4 = 1

x = -3

So at the point x = -3, we have a horizontal tangent.

If we wanted to solve for the actual point, just solve for y when x = -3.

y = [ ln(x + 4) ]^2

y = [ ln(-3 + 4) ]^2

y = [ ln(1) ]^2

y = [ 0 ]^2

y = 0

So at the point (-3, 0), we have a horizontal tangent.

Answer 2

the derivative tells you the slope of the tangent line to a curve at a specific point

being told that the tangent line is horizontal is equivalent to being asked to find where the derivative of this function is equal to zero

the derivative of this function is

2 ln[x+4]/(x+4)

this is zero when x=-3 since ln[1]=0

(it is also zero when x-> infinity but you probably don't care about that solution)