Question

Set up an equation of a tangent to the graph of the following function.
$y=3^x+3^{-2x}$ at the points with abscissa $x=1.$

Answer 1

If $x_1=1,y_1=3+3^{-2\left(1\right)}=\frac{28}{9}$

$\frac{dy}{dx}=(3^x-{2.3}^{-2x})\ln 3$

${\left(\frac{dy}{dx}\right)}_{x=1}=\frac{25\ln 3}{9}=m$

$(x-x_1)m=y-y_1$

$(x-1)\frac{25\ln 3}{9}=y-\frac{28}{9}$

$\left(25\ln 3\right)x-9y+28-25\ln 3=0$