Math 132, Self-­‐Improving-­‐Exercises 7, Instructor: Siyang Yang 1
Extreme values and derivatives:
(a) Identifying all the critical points of the curve y(x) = x 3 ! 2x 2
4
y'(x) = 3x 2 ! 4x = 0 " x = 0, x =
3
(b) Identifying all the maximum and minimum points (if they exist) of the curve
y(x) = x 3 ! 2x 2
# 4&
y"(x) = 6x ! 4 " y"(0) = !4 < 0 and y"% ( = 4 > 0 . Therefore, x=0 is local max, and x=4/3
$ 3'
is local minimum.
(c)
Use quotient rule to calculate derivative of function f (x) = csc x . Note that
1
.
csc x =
sin x
1
d
0 ! cos x
f '(x) = sin x =
= ! csc ( x ) " cot ( x )
dx
sin 2 x
(d) Use chain rule to calculate the derivative of the same function f (x) = csc x .
d ( sin x )
d ( sin x ) d ( sin x )
cos x
f '(x) =
=
"
= ! 2 = ! csc ( x ) " cot ( x )
dx
d ( sin x )
dx
sin x
!1
!1
Calculate derivative of function f (x) = cot x .
cos x
d
! sin 2 x ! cos 2 x
f '(x) = sin x =
= ! csc 2 x
dx
sin 2 x
(f)
Write out the definition of critical points, point of inflection and local
maximum.
See textbook.
(g) Find the critical points of f (x) = csc x + cot x .
(e)
f '(x) = ! csc x " cot x ! csc 2 x = 0
cos x
1
# ! 2 ! 2 = 0 # cos x = !1 # x = 2n$ + $
sin x sin x
(h) Does f (x) = csc x + 2 cot x have any local maximum values? Why?
2
No critical point or max or min, since f '(x) = ! csc x " cot x ! 2 csc 2 x = 0
.
cos x
2
# ! 2 ! 2 = 0 # cos x = !2 # No solution
sin x sin x
Mean Value Theorem:
(a) A marathon runner finishes the 26.2-mile New York City Marathon in exactly
2 hours. Please use mean value theorem to show that there must be at least
one instance when the runner was running at the speed of exactly 13.1
mile/hour (mph).
f (b) ! f (a) 26.2
The slope connecting the two ending points is
=
= 13.1 . According to
b!a
2
the mean value theorem, there exists at least one point at which the tangent is
equal to the slope.
(b) In the same Marathon, another runner finishes in 3 hours. With help of
mean value theorem, can you show that at least twice the runner was
Page 1 of 3 Math 132, Self-­‐Improving-­‐Exercises 7, Instructor: Siyang Yang 3
running at exactly 8 mile/hour (mph), assuming that the initial and final
speeds are zero.
f (b) ! f (a) 26.2
The slope connecting the two ending points is
=
= 8.73 . According to
b!a
3
the mean value theorem, there exists at least one point at which the tangent is
equal to the slope of 8.73 mph. In addition, since the speed increases from 0 and
finally finishes at 0, at least twice the runner was running at 8 mph.
Checking the Mean Value Theorem:
f (b) ! f (a)
Find the value or values of c that satisfy the equation f '(c) =
b!a
2/3
(a)
f (x) = x , [0,1]
Solution:
2
f '(x) = x !1/ 3
3
f (0) = 0
f (1) = 1
1! 0
" f '(c) =
=1
1! 0
2
" c !1/ 3 = 1
3
3
8
# 2&
"c=% ( =
$ 3'
27
(b)
f (x) = x ! 1 , [1, 3]
Solution:
1
f '(x) =
2 x !1
f (1) = 0
f (3) = 2
2!0
1
=
3!1
2
1
1
"
=
2 c !1
2
" f '(c) =
2
# 2&
3
"c=%
+1=
(
2
$ 2 '
(c)
#x3 , ! 2 " x " 0
CHALLENGING QUESTION what about f (x) = $ 2
? And what if
% x , 0<x " 2
#x , ! 2 " x " 0
f (x) = $ 2
% x , 0<x " 2
Solution:
First, you need to verify that f(x) is continuous and differentiable on the domain [2,2] in order for the mean value theorem to be applicable. Since x 2 and x 3 are
each differentiable, one needs to prove the function f(x) is continuous and
Page 2 of 3 Math 132, Self-­‐Improving-­‐Exercises 7, Instructor: Siyang Yang differentiable at the joining point x=0. Proof is as follows:
# x 3 =0 , ! 2 " x " 0
f (0) = $ 2
& f (x) is continuous at x = 0
x
=0
,
0<x
"
2
%
.
# 3x 2 =0 , ! 2 " x " 0
f '(0) = $
& f '(x) is differentiable at x = 0
% 2x=0 , 0<x " 2
Then, one can find the two values of c according to the MVT as follows:
3
f (!2) = ( !2 ) = !8
f (2) = ( 2 ) = 4
2
" f '(c) =
4 ! ( !8 )
=3
2 ! ( !2 )
If c < 0, then f '(c) = 3c 2 = 3 " c = !1
3
If c > 0, then f '(c) = 2c = 3 " c =
2
#x , ! 2 " x " 0
And for the second part f (x) = $ 2
, one can easily verify that this
% x , 0<x " 2
function is not differentiable at x=0 such that MVT is not applicable.
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