The graph of y = f ( x ) (solid) and its tangent line at x = 0 (dashed) is shown; a second point on the tangent line is also marked. Assume f ' ( x ) is continuous.

Question

I need help with this. I'm really confused and my professor is not very good at teaching.

Image Link-> https://imgur.com/a/XI3coTT

Answer 1

My guess is that the intent is for you to use L'Hospital's rule to answer these questions (although it is possible to answer part a) without doing that.

lim_x->0f(x)/x is an indeterminate form 0/0 so you can use the "rule". That limit is equivalent to:

lim_x->0f'(x)/1. Now how do we evaluate f'(x) as x -> 0? A formula for f'(x) will be the equation of the tangent line and as x->0 the result will be the slope of the tangent line at x = 0. So, calculate the slope of the tangent line by using the two points (0,0) and (3, -4). f'(0) = -4/3. That is the answer to part a).

For part b).

lim_x->0f(x)/(4e^x-4) is also an indeterminate form (0/0). So apply the "rule":

= lim_x->0f'(x)/4e^x = -4/3 / 4 = -4/3 (1/4) = -1/3.

You could also write the equation of the parabola and then find a formula for f'(x) and take the limit as x-> 0 by direct substitution.

The equation of the parabola is y = 2/9(x-3)²-2, so y' = 4/9(x-3) and f'(0) = 4/9(-3)= -4/3 -- the slope of the tangent line at x = 0!