140 STEP 4. Review the Knowledge You Need to Score High- A functionfis continuous on the interval
[−1, 4] withf(−1)=0 andf(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 whichf is
increasing or decreasing.
(b)Find where fhas 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 off.- Evaluate limx→π
2 x
sinx
.
- Evaluate lim
x→^12
3 − 6 x
4 x^2 − 1.
- Find the polar equation of the ellipse
x^2 + 4 y^2 =4.
7.9 Solutions to Practice Problems
Part A The use of a calculator is not
allowed.- Condition 1: Sincef(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) sincef′(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
, andf′′(x)=
−68(x^2 −25)^2 −2(x^2 −25)(2x)(− 68 x)
(x^2 −25)^4=
68(3x^2 +25)
(x^2 −25)^3.
Setf′′>0. Since (3x^2 +25)>0,
⇒(x^2 −25)^3 > 0 ⇒x^2 − 25 >0,
x<−5orx>5. Thus,f(x) is concave
upward on (−∞,−5) and (5,∞) and
concave downward on (−5, 5).- Step 1: f(x)=x+sinx,
f′(x)= 1 +cosx,
f′′=−sinx.
Step 2: Setf′′(x)= 0 ⇒−sinx=0or
x=0,π,2π.