Find all $x$ values such that the tangent line to the given curve satisfy a specific property

Question

Find all $x$ values such that the tangent line to the given curve satisfy a specific property.

(a) $y=\frac{x²+1}{x+2}$; horizontal. (Answer = $-2 +\sqrt3$)

(b)$y=\frac{x²+1}{x+1}$; parallel to $y=x$. (Answer = none)

The tangent problem is confusing me a bit. Could anybody walk me through the steps I need to solve this problem? Thanks in advance.

Answer 1

a) you want the line to be horizontal means parallel to the x-axis means the slope should be 0 thus $$\frac{dy}{dx}=0$$ you have $$y=\frac{x²+1}{x+2}$$ thus $$\frac{dy}{dx}=\frac{2x(x+2)-(x^2+1)}{(x+2)^2}=\frac{(x^2+4x-1)}{(x+2)^2}$$

$\frac{dy}{dx}=0$ gives $x^2+4x-1=0$ solve for x

b) in the second part you want the line to be parallel of $y=x$ thus the slope must be 1 thus $$\frac{dy}{dx}=1$$ $$\frac{(x^2+4x-1)}{(x+2)^2}=1$$ $$x^2+4x-1=x^2+4x+4$$ $$-1=4$$ which is not true thus you can not draw a tangent to the given curve y that is parallel to $y=x$

Answer 2

Hint: For a) $$f'(x)=\frac{x^2+2x-1}{(1+x)^2}$$ so it must be $$f'(x)=0$$ For b) such Tangent line must have the form

$y=x+n$ so solve the equation $$x+n=\frac{x^2+1}{x+1}$$ and set the discriminante equal to Zero.