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Let $h$ be the function of domain $\mathbb{R}\setminus[0]$

$h(x) = \left\{ \begin{array}{lcc} e^x, & x < 0 \\ \\ \ln(x+1), & x > 0 \\ \\ \end{array} \right.$

For a certain real number $b$, the equation $y = b$ intercepts the graph of $h$ in 2 points, one with a negative x coordinate and one with a positive one that will be designated by A and B, respectively. It is known that $\overline{AB}=2.8$

Find $b$ and the x coordinates of A and B. Use your calculator for this.

To solve this I tried:

$$X_A < 0 \\ \\ Y_A = e^{X_A}\\ X_B > 0 \\ Y_B = \ln(X_B +1 ) \\ Y_A = Y_B = b$$

I also know that $b$ is in the interval $]0;1[$ and $X_A \in ]-\infty;0[ $ and $X_B \in ]0;+\infty[$.

$$\\ 2.8 = \sqrt{(X_A-X_B)^2+(Y_A-Y_B)^2} \Leftrightarrow \\ 2.8 = \sqrt{(X_A-X_B)^2+ 0} \Leftrightarrow \\ 2.8 = X_A-X_B$$

I don't think this is correct though, because $2.8 + X_B = X_A$ with $X_A$ being a negative value seems impossible.

I think I might have swapped $X_A$ with $X_B$. In that case, $2.8 = X_B-X_A$ becomes $2.8 + X_A = X_B$ which could work. Is this correct?

What did I do wrong? How do I solve this?

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    You need to choose the solution with $X_B>X_A$. Think of $\sqrt{(-1)^2}$. The answer is 1, not -1 – 2017-02-13

1 Answers 1

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I solved it!

I realised I had swapped $X_A$ with $X_B$ in the distance formula ($X_B$, being the second value, should be first), and with that I could solve it. Here it is:

$$2.8 = X_B-X_A$$

Then I tried isolating $X_A$ and $X_B$ in their respective functions:

$$e^{X_A} = b \Leftrightarrow X_A = \ln(b)$$ $$\ln(X_B+1) = b \Leftrightarrow e^b = X_B+1 \Leftrightarrow e^b-1 = X_B$$

Then I replaced those values in the formula:

$$2.8 = X_B-X_A \Leftrightarrow 2.8 = e^b-1-\ln(b)$$

I don't think that can be solved analitically so that's probabily why my book requires the calculator to be used.

So I just have to put that in the calculator and calculate the interception. There are several interceptions, an example would be at $b \approx .065$ which my book says is correct.