Question

show that the curves y=e^3x and y =3e^x each have the same gradient where they cross the y axis. Find the equation of the tangents to each graph at these points

Answer 1

In general, for a function f which is differentiable at a point x=a, the equation of the tangent to the graph of f at that point is given by
\[ y = f'(a)(x-a) + f(a) \ \ \ -- (*)\]
This makes complete sense because in the limit as \( x \rightarrow a \),
\[ \frac{ f(x) - f(a)}{x-a} \longrightarrow f'(a) \]

Use the formula (*) for your two functions at x = a = 0.

Answer 2

what does --(*) mean?

Answer 3

I mean I'm labeling or calling that formula (*), so I can reference it later. It's not a mathematical function.

Answer 4

oh i see
hehehe

Answer 5

so for y=e^3x would the eqn of the tangent be
y= (3e^3x )(x-1) + e^3x?