Solution of following equation

Asked Jun 17 '24 at 17:58

Active Jun 17 '24 at 18:07

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i'm in trouble with the following problem:

Find all values of $k$ give that the following equation has exactly $2$ solutions: $$ (x^3-3)e^x = k, \quad k \in \mathbb{R}$$

I have that $f'(x) = e^x(x^3+3x^2-3)$. If I put this derivative equal to $0$ I don't know how to procede.

If I use Wolframalpha i see that the equation $x^3+3x^2-3=0$ has $3$ solutions, but using wolframalpha again with the the $f(x)$ original it says that we have just one minimum. I don't understand. Could someone help me please? Thanks

edited Jun 17 '24 at 18:07

asked Jun 17 '24 at 17:58

Skills

1,483

Did you mean “find all values of k”? Because if x is a given number, then the left hand side of the equation would simply be a fixed number, and there would be nothing to solve. It would look something like $123=k$. – Divide1918 Jun 17 '24 at 18:04
@Divide1918 correct. thanks! – Skills Jun 17 '24 at 18:07
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Try finding intervals where $g(x)=(x^3-3)e^x$ is increasing/decreasing and it's asymptotes – Dunham Jun 17 '24 at 18:17
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Look closer at $f$'s plot. It doesn't have "just one minimum". – Anne Bauval Jun 17 '24 at 19:30
@AnneBauval thanks. But i can answer to the question? – Skills Jun 17 '24 at 19:38
@Dunham how? If i can't find the zeros of the derivative? – Skills Jun 17 '24 at 19:39
You can find the zeros of a cubic using Cardano's method. – Dunham Jun 17 '24 at 20:30
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Does this answer your question? An interesting Cubic equation $x^3-3x-1=0$ Of course, first make the substitution $x=y-1$ in your equation $x^3+3x^2-3=0$. – Anne Bauval Jun 18 '24 at 00:25

Solution of following equation

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