Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.2 THE MEAN VALUE THEOREM ¤ 153
3. f(x) = √ x − 1 3
x, [0, 9]. f, being the difference of a root function and a polynomial, is continuous and differentiable
on [0, ∞), so it is continuous on [0, 9] and differentiable on (0, 9). Also,f(0) = 0 = f(9). f 0 (c) =0
⇔
1
2 √ c − 1 3 =0 ⇔ 2 √ c =3 ⇔ √ c = 3 2
conclusion of Rolle’s Theorem.
⇒
c = 9 4 , which is in the open interval (0, 9),soc = 9 satisfies the
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5. f(x) =1− x 2/3 . f(−1) = 1 − (−1) 2/3 =1− 1=0=f(1). f 0 (x) =− 2 3 x−1/3 ,sof 0 (c) =0has no solution. This
does not contradict Rolle’s Theorem, since f 0 (0) does not exist, and so f is not differentiable on (−1, 1).
7.
f(8) − f(0)
8 − 0
= 6 − 4
8
= 1 4 . The values of c which satisfy f 0 (c) = 1 seem to be about c =0.8, 3.2, 4.4,and6.1.
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9. (a), (b) The equation of the secant line is
(c) f(x) =x +4/x ⇒ f 0 (x) =1− 4/x 2 .
y − 5= 8.5 − 5
8 − 1 (x − 1) ⇔ y = 1 x + 9 . So f 0 (c) = 1 2
⇒ c 2 =8 ⇒ c =2 √ 2,and
2 2
f(c) =2 √ 2+ 4
2 √ =3√ 2. Thus, an equation of the
2
tangent line is y − 3 √ √
2= 1 2 x − 2 2 ⇔
y = 1 2 x +2√ 2.
11. f(x) =3x 2 +2x +5, [−1, 1]. f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous
and differentiable on R.
c =0, which is in (−1, 1).
f 0 (c) =
f(b) − f(a)
b − a
⇔
6c +2=
f(1) − f(−1)
1 − (−1)
= 10 − 6
2
=2 ⇔ 6c =0 ⇔
13. f(x) =e −2x , [0, 3]. f is continuous and differentiable on R, so it is continuous on [0, 3] and differentiable on (0, 3).
f 0 f(b) − f(a)
(c) = ⇔ −2e −2c = e−6 − e 0
⇔ e −2c = 1 − e−6
1 − e
−6
⇔ −2c =ln
⇔
b − a
3 − 0
6
6
c = − 1 1 − e
−6
2 ln ≈ 0.897, which is in (0, 3).
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