MATH 1A SECTION: OCTOBER 28, 2013
Recall Rolle’s Theorem and the Mean Value Theorem.
Theorem 1 (Rolle’s Theorem). Let f be a real-valued function, continuous on [a, b] (a < b)
and differentiable on (a, b). Suppose that f (a) = f (b). Then there exists c in (a, b) such that
f 0 (c) = 0.
Theorem 2 (Mean Value Theorem). Let f be a real-valued function, continuous on [a, b]
(a < b) and differentiable on (a, b). Then there exists c in (a, b) such that
f (b) − f (a)
.
f 0 (c) =
b−a
Think of the mean value theorem as a tilted version of Rolle’s Theorem; this also happens
to be how the proof goes.
1. If f is differentiable, f (1) = 10, and f 0 (x) ≥ 2 for 1 ≤ x ≤ 4, how small can f (4) be?
2. Does there exist a differentiable function f such that f (0) = −1, f (2) = 4, and f 0 (x) ≤ 2
for all x?
√
3. Find the point on the curve y = x that is closest to the point (3, 0).
].
4. Find the absolute maximum value of f (t) = t + cot( 2t ) on [ π2 , 3π
2
Solutions and Commentary
1. Intuitively, we should think that if f 0 (x) = 2, we would have f (x) = 2x+8, and f (4) = 16.
If f 0 (x) ≥ 2, the function can only go up more quickly, and we should have f (4) ≥ 16.
That’s the right intuition, but to formalize it, we should use the Mean Value Theorem.
Since our function is differentiable, we can apply the Mean Value Theorem. By the
Mean Value Theorem, we have that there exists some c in (1, 4) such that
f (4) − 10
f (4) − f (1)
=
.
f 0 (c) =
4−1
3
Then f (4) − 10 = 3f 0 (c), so f (4) = 3f 0 (c) + 10. Since f 0 (x) ≥ 2 for all x in [1, 4], we see
that f (4) = 3f 0 (c) + 10 ≥ 3 · 2 + 10 = 16. This is what we wanted.
2. Suppose we have a differentiable function f such that f (0) = −1 and f (2) = 4. Then,
since the function is differentiable, we can apply the Mean Value Theorem. This means
that there exists c in (0, 2) such that
f (2) − f (0)
4 − (−1)
5
f 0 (c) =
=
= > 2.
2−0
2−0
2
0
Therefore, we are forced to have f (x) > 2 somewhere in the interval (0, 2); the answer to
the question is no.
3. This is very much like a point that we did√last time.
Let (x, y) be a point on the curve y = x. We are interested in the distance between
(x, y) and (3, 0). To find this distance, use the distance formula:
p
p
d = (x − 3)2 + (y − 0)2 = (x − 3)2 + y 2 .
We need to minimize this, but we have two √
variables, and we don’t like this. But recall
that the point (x, y) is on the curve, so y = x and hence y 2 = x. Plugging in, we have
p
d = (x − 3)2 + x.
It suffices to minimize d2 rather than d; after all, to minimize d, we can minimize its
square, and then take the square root.
So we need to find the minimum value of f (x) = (x − 3)2 + x. We have f 0 (x) =
2(x − 3) + 1 = 2x − 5. The critical number happens at f 0 (x) = 0, which gives 2x = 5 or
x = 25 . Since f 00 ( 52 ) = 2 > 0, we see that we indeed have a minimum at x = 52 , so this
q gives our closest point as 25 , 52 .
4. To do this, we differentiate and find critical numbers. We have f 0 (t) = 1 − 21 csc2 ( 2t ).
Solving f 0 (t)
= 0, we have 1 = 12 csc2 ( 2t ), so csc2 ( 2t ) = 2, and hence sin2 ( 2t ) = 12 , so
√
sin( 2t ) = ± 22 . This means that we can have 2t = π4 , 3π
, . . . , so therefore t = π2 , 3π
,....
4
2
There are no critical numbers in the interior of the interval, and we only need to consider
the endpoints!
We check the endpoints to see that f ( π2 ) = π2 + 1 and f ( 3π
) = 3π
− 1. Therefore,
2
2
3π
3π
f ( 2 ) = 2 − 1 is the absolute maximum.
2