1. Tangent lines

I have a problem that I do not know how to start. If anyone could give me a hint on how to start it, that would be great.

Let $f$ be the function given by $f(x) = e^{-x}$, and let $g$ be the function given by $g(x) = kx$, where $k$ is the nonzero constant such that the graph of $f$ is tangent to the graph of $g$.

Find the x-coordinate of the point of tangency and the value of $k$.

2. By condition of tangance df/dx x dg/dx = -1

so $-e^{-x} * k = -1
$

$e^{-x} = 1/k$

$e^x = k$
$x = \ln k$

So g(x) is tangent to f(x) at point $x = \ln k$

Now for x = \ln k
you cordinates of both curve should be same

so x = \ln k satisfies
$e^{-x}= kx$

solve for k

3. Tangent

Hello Cursed
Originally Posted by Cursed
I have a problem that I do not know how to start. If anyone could give me a hint on how to start it, that would be great.

Let $f$ be the function given by $f(x) = e^{-x}$, and let $g$ be the function given by $g(x) = kx$, where $k$ is the nonzero constant such that the graph of $f$ is tangent to the graph of $g$.

Find the x-coordinate of the point of tangency and the value of $k$.
First, you need to find the value of $x$ where the two graphs meet: that is where $f(x) = g(x)$, or

$e^{-x}= kx$ (1)

Then if $g(x)$ is a tangent to $f(x)$ for this value of $x$, their gradients are equal at this point. So $f'(x) = g'(x)$ or:

$-e^{-x} = k$

$\Rightarrow e^{-x} = -k$ (2)

Now you can use (2) to eliminate the $e^{-x}$ term in (1), and solve the resulting equation for $x$. This is the $x$-coordinate of the point of tangency, which you can then use to find the value of $k$, by substituting it back into (1).

Can you complete it now?