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I have the function $$y = \frac{(x-a)e^x}{(x-b)}$$ and I am told that the curve has stationary points under the following conditions -

$$a-b < 0 \quad\text{or}\quad a-b>4$$

I started by differentiating the equation to get $$ y'=\frac{(e^x+xe^x-ae^x)(x-b)-xe^x+ae^x}{(x-b)^2} \tag{1}$$ then simplified to get $$y'=\frac{e^x(x^2-x(a+b)+(a-b+ab))}{(x-b)^2} \tag{2}$$

I then set this equal to zero and attempted to use $b^2-4ac > 0$ on the quadratic. However, I am not really sure if this is the correct way to go about it, as I could not see a solution from this.

Any and all help is appreciated.

Blue
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AH_01
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1 Answers1

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Hint: $$ (a+b)^2-4(a-b+ab) = (a-b)^2-4(a-b)$$

ir7
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  • thank you, makes sense now. – AH_01 Jul 19 '20 at 19:07
  • @AH_01 I'm glad it helps. Unless you are waiting for other answers, please consider marking this answer as accepted (marking procedure available here: https://stackoverflow.com/help/someone-answers ) – ir7 Jul 19 '20 at 19:13