Interpreting derivative challenge

Let f be a differentiable function for all x. If f of negative 2 is equal to 3 and f prime of x is less than or equal to 7 for all x, then what is the largest possible value of f of 10? And so I encourage you to think about this on your own, pause the video, try to figure out the largest possible value for f of 10. And then we'll work through it together. So I'm assuming you've given a go at it. So let's visualize this. So let me draw some axes here. So let's say that's my x-axis. That's my x-axis right over there. And this right over here is my y-axis. That's my y-axis [INAUDIBLE] I'll graph y equals f of x. And they tell us f of negative 2 is equal to 3. And the two axes aren't going to be drawn to scale. So let's say this is negative 2. And this right over here is the point negative 2 comma 3. And they tell us that f prime of x is a less than or equal to 7, that the instantaneous slope is always less than or equal to 7. So really, the way to get the largest possible value of f-- we don't have to necessarily invoke the mean value theorem, although the mean value theorem will help us know for sure-- is to say well, look, the largest possible value of f of 10 is essentially if we max this thing out. If we assume that the instantaneous rate of change just stays at the ceiling right at 7. So if we assumed that our function, the fastest growing function here would be a line that has a slope exactly equal to 7. So the slope of 7 would look-- and obviously, I'm not drawing this to scale. Visually, this looks more like a slope of 1, but we'll just assume this is a slope of 7 because it's not at the same-- the x and y are not at the same scale. So slope is equal to 7. And so if our slope is equal to 7, where do we get to when x is equal to 10? When x is equal to 10, which is right over here, well what's our change in x? So what's our change in x? Let's just think about it this way. Our change in y over change in x is going to be what? Well our change in y is going to be f of 10 minus f of 2. f of 2 is 3, so minus 3, over our change in x. Our change in x is 10 minus negative 2. 10 minus negative 2 is going to be equal to 7. This is the way to max out what our value of f of 10 might be. If at any point the slope were anything less than that, because remember, the instantaneous rate of change can never be more than that. So if we start off even a little bit lower, than the best we can do is get to that. Remember, we can't do something like that. That would get us too steep. So it has to be like that. And then we would get to a lower f of 10. Every time you have a slightly lower rate of change, then it kind of limits what happens to you. So remember, our slope can never be more than 7. So this part should be parallel. So this should be parallel to that right over there. This should be parallel. But we can never have a higher slope than that. So the way to max it out is to actually have a slope of 7. And so what is f of 10 going to be? So let's see, 10 minus negative 2, that is 12. Multiply both sides by 12, you get 84. So f of 10 minus 3 is going to be equal to 84. Or f of 10 is going to be equal to 87. So if you have a slope of 7, the whole way, you travel 12. That means you're going to increase by 84. If you started at 3, you increase by 84, you're going to get to 87.