Jack's question at Yahoo! Answers regarding finding gradients

[MarkFL]

Here is the question:

LOGARITHMIC FUNCTION, MATHS QUESTION PLEASEE HELPP ME IM BEGGIN YOU!!!?

Consider the curve f(x)=ln(x+1). find the gradients of the possible tangents to f(x) which makes of 45 degrees with the tangent of f(x) at the point where x=1

Please explain your answer thankyou

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Jack,

First, we want to find the gradient of the line tangent to the given logarithmic curve where $x=1$. To do so, we must differentiate the curve with respect to $x$:

\(\displaystyle f'(x)=\frac{1}{x+1}\)

Now, we evaluate this for $x=1$:

\(\displaystyle f'(1)=\frac{1}{2}\)

Then, to find the possible gradients $m$ that make an angle of 45° with the tangent line at its point of tangency, we may equate the magnitude of the difference in the angles of inclination to \(\displaystyle \frac{\pi}{4}\).

\(\displaystyle \left|\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right|=\frac{\pi}{4}\)

\(\displaystyle \tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right)=\pm\frac{\pi}{4}\)

Taking the tangent of both sides, we find:

\(\displaystyle \tan\left(\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right)=\tan\left(\pm\frac{\pi}{4} \right)\)

Using the angle-difference identity for tangent on the left, and simplifying the right, we obtain:

\(\displaystyle \frac{m-\frac{1}{2}}{1+\frac{m}{2}}=\pm1\)

Now we may solve for $m$. Multiply through by \(\displaystyle 2\left(1+\frac{m}{2} \right)\)

\(\displaystyle 2m-1=\pm(2+m)\)

Square both sides, then arrange as the difference of squares:

\(\displaystyle (2m-1)^2-(2+m)^2=0\)

Apply the difference of squares formula:

\(\displaystyle (2m-1+2+m)(2m-1-2-m)=0\)

Combine like terms:

\(\displaystyle (3m+1)(m-3)=0\)

Hence the possible gradients are:

\(\displaystyle m=-\frac{1}{3},3\)

As we should expect, the product of the two gradients is -1 as the two lines would be perpendicular to one another.