Find the normal to the function

Question

Good day everyone, I think my teacher had made a mistake in a answer. She claims that this function cannot have normal that is parallel to $y=2x+7$

$f(x)=\ln(x+1)$

In my calculations it has : $y=-\frac{1}{2}x -\frac{1}{4} + \ln\frac{1}{2} $

Can you ansure me that it has ?

Have you tried sketching a diagram? Always useful when geometric things are in doubt. — Mark Bennet, Dec 26 '16 at 11:43
You are looking for a normal parallel to the given line, not perpendicular to it, which would make the tangent parallel to the given line. — Mark Bennet, Dec 26 '16 at 11:45

score 1 · Accepted Answer · answered Dec 26 '16 at 11:37

1

Your teacher is correct.

The derivative of the function $f(x)$ is $f'(x)=\frac{1}{x+1}, x\geq -1$. The normal to $y=2x+7$ has a derivative of $-\frac{1}{2}$.

Since $x\geq -1$, there is no solution to $x$ which makes $f'(x)=\frac{1}{x+1}=-\frac{1}{2}$.

answered Dec 26 '16 at 11:37

projectilemotion

1 Answers1