1. ## Confused about TLA (I think)

Suppose that f has a continuous second derivative for all x, and that f(0) = 1, f '(0) = 2,
and f ''(0) = 0.
Does f have an inflection point at x = 0? Explain your answer.
Let g'(x) = (3(x^2) + 2)f(x) + (x^3 + 2x + 5)f '(x). The point (0, 5) is on the graph of g. Write the
equation of the tangent line to g at this point.
Use your tangent line to approximate g(0, 3).
Find g''(0)

I'm just not sure where to even begin with this problem. I have never had a problem like this and it seems so overwhelmingly complicated. If anyone wants to help me out, it would be much appreciated.

I think the answer to the first part is yes, because the second derivative indicates inflection points or changes in concavity. So if it is at zero, then it has to be changing from concave up to concave down or concave down to up...

2. My mnemonic device for the Second Derivative Test (note what happens in the zero case):

It could be concave up, or concave down, or a point of inflection. Pretty confused individual! All this to say, I don't think you can know if you have a point of inflection there or not. The function $\displaystyle x^{4}$ has its second derivative zero at the origin, but the origin is not a point of inflection: it's concave up there. Similarly, $\displaystyle -x^{4}$ is concave down at the origin - no point of inflection. Finally, $\displaystyle x^{3}$ does have a point of inflection at the origin.

3. In order that there be an inflection point at a given x value, the second derivative must change sign and so, assuming it is continuous, must be 0 there. However, the converse is not true. A function (the second derivative) may be 0 without passing from positive to negative or vice versa. For example, the function $\displaystyle f(x)= x^4$ has second derivative $\displaystyle 12x^2$ which is 0 at x= 0 but is positive for all other x. That function does not have an inflection point at x= 0 even though it second derivative is 0 there.

4. thanks for the help on the first part! That makes perfect sense now. Any hints on the second and third part of the problem? I understand that I need to take the derivative of g(x) to get the slope and use the point (0,5) to find B. But how am I supposed to do that when there is an f(x) and an f''(x) in g(x)?

5. Originally Posted by rogers8315
Suppose that f has a continuous second derivative for all x, and that f(0) = 1, f '(0) = 2,
and f ''(0) = 0.
You know all of these values of f they are in your first post!