Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. Let y = f(x) =e −x2 . The area of the rectangle under the curve from −x to x is A(x) =2xe −x2 where x ≥ 0. We maximize
A(x): A 0 (x) =2e −x2 − 4x 2 e −x2 =2e −x2 1 − 2x 2 =0 ⇒ x = 1 √
2
. This gives a maximum since A 0 (x) > 0
for 0 ≤ x< 1 √
2
and A 0 (x) < 0 for x>
1 √
2
. We next determine the points of inflection of f(x). Notice that
f 0 (x) =−2xe −x2 = −A(x). Sof 00 (x) =−A 0 (x) and hence, f 00 (x) < 0 for − 1 √
2
0 for x 1 √
2
.Sof(x) changes concavity at x = ± 1 √
2
, and the two vertices of the rectangle of largest area are at the inflection
points.
3. First, we recognize some symmetry in the inequality: ex + y
xy
≥ e 2 ⇔ ex x · ey
y
≥ e · e. This suggests that we need to show
that ex x
≥ e for x>0. If we can do this, then the inequality
ey
y ≥ e is true, and the given inequality follows. f(x) =ex x
⇒
f 0 (x) = xex − e x
x 2 = ex (x − 1)
x 2 =0 ⇒ x =1. By the First Derivative Test, we have a minimum of f(1) = e, so
e x /x ≥ e for all x.
ax 2 +sinbx +sincx +sindx
5. Let L =lim
.NowL has the indeterminate form of type 0 , so we can apply l’Hospital’s
x→0 3x 2 +5x 4 +7x 6
0
2ax + b cos bx + c cos cx + d cos dx
Rule. L =lim
. The denominator approaches 0 as x → 0, so the numerator must also
x→0 6x +20x 3 +42x 5
approach 0 (because the limit exists). But the numerator approaches 0+b + c + d,sob + c + d =0. Apply l’Hospital’s Rule
2a − b 2 sin bx − c 2 sin cx − d 2 sin dx
again. L =lim
= 2a − 0
x→0 6+60x 2 +210x 4 6+0 = 2a 6 , which must equal 8. 2a
=8 ⇒ a =24.
6
Thus, a + b + c + d = a +(b + c + d) =24+0=24.
7. Differentiating x 2 + xy + y 2 =12implicitly with respect to x gives 2x + y + x dy dy dy
+2y =0,so
dx dx dx = − 2x + y
x +2y .
At a highest or lowest point, dy =0 ⇔ y = −2x. Substituting −2x for y in the original equation gives
dx
x 2 + x(−2x)+(−2x) 2 =12,so3x 2 =12and x = ±2. Ifx =2,theny = −2x = −4, andifx = −2 then y =4.
Thus, the highest and lowest points are (−2, 4) and (2, −4).
9. y = x 2 ⇒ y 0 =2x, so the slope of the tangent line at P (a, a 2 ) is 2a and the slope of the normal line is − 1 for a 6= 0.
2a
An equation of the normal line is y − a 2 = − 1
2a (x − a). Substitute x2 for y to find the x-coordinates of the two points of
intersection of the parabola and the normal line. x 2 − a 2 = − x 2a + 1 2
⇒ 2ax 2 + x − 2a 3 − a =0 ⇒
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