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I am not sure how to find points of $y=\frac{x}{2}+\frac{1}{2x+4}$ where the slope is $\frac{-3}{2}$ without looking at a graph.

I can simplify the function by writing it as $y=\frac{x(x+2)+1}{2(x+2)}$, but I have no idea what to look for to discover the function.

I think that derivatives are equal to the slope of the line at a certain point, but I can't check every point in the function.

How can I find the points where the slope is $\frac{-3}{2}$?

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Hint: Solve the equation $$f'(x)=-\frac{3}{2}$$ for $$x$$

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  • $\begingroup$ I get it! Thank you! $\endgroup$ – LuminousNutria May 10 at 18:47
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    $\begingroup$ What have you got? $\endgroup$ – Dr. Sonnhard Graubner May 10 at 18:50
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    $\begingroup$ It is $$x_1=-\frac{5}{2}$$ or $$x_2=-\frac{3}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner May 10 at 18:51
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I realize you didn't want a graph, but here: you can check your answer:

enter image description here

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I realise you may know the answer but just for completeness sake, we have $y=\frac{x}{2}+\frac{1}{2x+4}$, as you said! Then differentiating this we get: $$ y'(x)=\frac{1}{2}-\frac{1}{2(x+2)^2}$$ which is a formula for the gradient! We want the gradient at $\frac{-3}{2}$, so we set $y'(x)=\frac{-3}{2}$, which means: $$\frac{-3}{2}=\frac{1}{2}-\frac{1}{2(x+2)^2}$$

Then solving this we get the quadratic $(x+2)^2=\frac{1}{4}$, so solving this we get $x=\frac{-5}{2}$ and $x=\frac{-3}{2}$. If you don't understand the simplification which I skipped just ask :)

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