Find all values of c, if any so that the tangent line to the graph of f at x=c passed through (0,0)

Question

Let 𝑓𝑥=(𝑒^x)−𝑥, and let 𝑐 be a real number. Find 𝑓′(𝑥) and all values of 𝑐, if any, so that the tangent line to the graph of 𝑓 at 𝑥=𝑐 passes through the point (0,0). If no such 𝑐 value exists, explain why.

Hello, i'm having quite some trouble processing what the values of c are, and how i would go about finding them, and figuring out which ones don't have any c values. Any help is appreciated!

Heres also a similar problem that asks for the same thing:

g(x) = x-lnx

Answer 1

So f'(x) = e^x - 1 which is e^c - 1 for x = c

The equation of the the line passing through x = c is

y - f(c) = f^'(c) (x-c) point slope form with slope f'(c) passing through point (c,f(c))

y - e^c + c = (e^c-1)(x-c)

y = (e^c-1)x - ce^c + c +e^c - c

y = ()x + (1-c)e^c Since this is in slope/intercept form, you can pick off the values of c that make the y intercept 0 (1-c) = 0 c = 1

If the intercept term was e^c only, there would be no c for the tangent to go through the intercept