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MarkFL

Staff member
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

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

MarkFL

Staff member
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:

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

$$\displaystyle m=-\frac{1}{3},3$$