Question

Set up an equation of a tangent to the graph of the following function.
$y=\frac{1}{2}(e^{x/2}+e^{-x/2})$ at the point which abscissa $x=2\ln 2.$

Answer 1

$y=\frac{1}{2}(e^{x/2}+e^{-x/2})$

The y-coordinate of the point with abscissa =2ln2

$y=\frac{1}{2}(e^{2\ln 2/2}+e^{-2\ln 2/2})=\frac{1}{2}(2-2)=0$

The point is $(2\ln 2,0)$

$\frac{dy}{dx}=\frac{1}{4}(e^{x/2}-e^{-x/2})$

${\left(\frac{dy}{dx}\right)}_{x=2\ln 2}=\frac{1}{4}\times 4=1$

The equation of tangent passing through (2ln2,0) and having slope m=1 is

$1(x-2\ln 2)=y-0$

$x-y-2\ln 2=0$