MA 3972-MA-Book April 11, 2018 17:21Graphs of Functions and Derivatives 1678.7 Cumulative Review Problems
(Calculator) indicates that calculators are
permitted.
- Find
dy
dx
if (x^2 +y^2 )^2 = 10 xy. - Evaluate limx→ 0
√
x+ 9 − 3
x.
- Find
d^2 y
dx^2
ify=cos(2x)+ 3 x^2 −1. - (Calculator) Determine the value ofksuch
that the function
f(x)={
x^2 −1, x≤ 1
2 x+k, x> 1is continuousfor all real numbers.- A functionf is continuous on the interval
[−1, 4] withf(−1)=0 and f(4)=2 and
the following properties:
INTERVALS (−1, 0) X= 0 (0, 2)X=2 (2, 4)
f′ + undefined + 0 −
f′′ + undefined − 0 −(a) Find the intervals on which fis
increasing or decreasing.
(b) Find wheref has its absolute extrema.
(c) Find wheref has points of inflection.
(d) Find intervals on whichf is concave
upward or downward.
(e) Sketch a possible graph of f.8.8 Solutions to Practice Problems
Part A The use of a calculator is not
allowed.- Condition 1: Since f(x) is a polynomial, it
is continuous on [−1, 2].
Condition 2: Also,f(x) is differentiable
on (−1, 2) because f′(x)= 3 x^2 − 2 x−2is
defined for all numbers in [−1, 2].
Condition 3:f(−1)= f(2)=0. Thus,
f(x) satisfies the hypotheses of Rolle’s
Theorem, which means there exists acin
[−1, 2] such thatf′(c)=0. Set
f′(x)= 3 x^2 − 2 x− 2 =0. Solve
3 x^2 − 2 x− 2 =0, using the quadratic
formula and obtainx=
1 ±
√
7
3
.Thus,
x≈ 1 .215 or− 0 .549 and both values are
in the interval (−1, 2). Therefore,
c=1 ±
√
7
3.
- Condition 1:f(x)=exis continuous on
[0, 1].
Condition 2:f(x) is differentiable on
(0, 1) since f′(x)=exis defined for all
numbers in [0, 1].
Thus, there exists a numbercin [0, 1]
such that f′(c)=
e^1 −e^0
1 − 0
=(e−1).
Setf′(x)=ex=(e−1). Thus,
ex=(e−1). Take ln of both sides.
ln(ex)=ln(e−1)⇒x=ln(e−1).
Thus,x≈ 0 .541, which is in the interval
(0, 1). Therefore,c=ln(e−1). - f(x)=
x^2 + 9
x^2 − 25
,
f′(x)=
2 x(x^2 −25)−(2x)(x^2 +9)
(x^2 −25)^2=
− 68 x
(x^2 −25)^2
, and