Is there a value $x_0$ ∈ $\Bbb{R}$ such that the tangent to the graph of (x) = sinh(x) at $x_0$ is parallel to the line y = 2x−2? What about a tangent parallel to the line y = $\dfrac{1}{2}$x + 3? Draw a qualitative sketch showing all the solutions you find.
I know that sinh(x) = $\dfrac{1}{2}(e^x − e^{-x})$ and that the derivative of sinh(x) = $\dfrac{1}{2}(e^{-x} + e^x)$. The slopes of y = 2x−2 and y = $\dfrac{1}{2}$x + 3 are 2 and $\dfrac{1}{2}$, respectively. I then make $\dfrac{1}{2}(e^{-x} + e^x)$ = 2, $\dfrac{1}{2}$ in order to find a value that allows for the tangent of (x) to be parallel to the aforementioned functions. However, I am not sure how to compute $\dfrac{1}{2}(e^{-x} + e^x) = 2$, $\dfrac{1}{2}$ to get the necessary values.
Let $y=e^x$. Then you want to solve $y+\dfrac1y=4$.
Multiply through by $y\ne0$, and it's a quadratic equation:
$y^2+1=4y$, or $y^2-4y+1=0$.
Now solve the quadratic equation for $y$ and take $x=\ln y$.