 In this video, we provide the solution to question number 11 for the practice final exam for math 1210, in which case we're told that if a function f, it's first derivative, it's second derivative, are all continuous on their domains, and if f of two equals zero, f prime of two equals negative two and f double prime of two equals three, what can we say about the function happening at x equals two? So let's see what information we have. Since the function at two is equal to zero, this tells us that x equals two is an x intercept of the graph. That might be useful, maybe that's what they're asking about. If we look at f prime at two, since it's negative two, the derivative tells you the monotonicity of the function. So when the derivative is positive, the function's increasing. When the derivative is negative, that means the function's decreasing. And so we can tell that the function is decreasing at the value f equals two, excuse me, x equals two. And then finally, the second derivative measures the concavity of the function. So if the second derivative at two is equal to three, that tells us that the function must be concave upward at x equals two. A positive second derivative would be concave up, a negative second derivative would be concave down. So let's see if we find a result that says anything about that. So we're actually gonna select choice E right here because the function's decreasing in concave up at two. These other statements, we know that the function's not increasing so it can't be choices C and D. We know that the function's concave up so it couldn't be choice F or D. What do we know about, could it be a local minimum or maximum? Well, if it was a local maximum or local minimum, it would be an extremum. And that would require that the first derivative be zero at x equals two, which is not the case. So this is not an extremum. So the only choice that would be correct would be choice E.