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# Interpreting derivative challenge

Video transcript

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.