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Answer Key for Exam 2
1(a) lim x→ 1 x^4 + 6x^3 − 22 x + 15 x^4 + x^3 − 3 x^2 − x + 2
, so we have to do something. There must be at least one factor of x−1 in the top and in the bottom, but factoring is a less attractive option than L’Hˆopital’s rule in this case:
xlim→ 1
x^4 + 6x^3 − 22 x + 15 x^4 + x^3 − 3 x^2 − x + 2
LH = lim x→ 1
4 x^3 + 18x^2 − 22 4 x^3 + 3x^2 − 6 x − 1 still
LH = lim x→ 1
12 x^2 + 36x 12 x^2 + 6x − 6 =
Since we had to use L’Hˆopital’s rule twice, the top and bottom must each have had two factors of x − 1. In fact,
lim x→ 1
x^4 + 6x^3 − 22 x + 15 x^4 + x^3 − 3 x^2 − x + 2 = lim x→ 1
(x − 1)^2 (x + 3) (x + 5) (x − 1)^2 (x + 1) (x + 2)
= lim x→ 1
(x + 3) (x + 5) (x + 1) (x + 2)
=^4 ·^6
1(b) (^) xlim→∞ x^2 ex^2
, so we have to do something. By L’Hˆopital’s rule we have
xlim→∞^ x
2 ex^2
LH = lim x→∞
2 x 2 x ex^2 = (^) xlim→∞^1 ex^2
- Take the derivative with respect to x on both sides of x^7 + y^7 = 14x^2 y^4 :
d dx x^7 + d dx y^7 = d dx 14 x^2 y^4 = 14 d dx x^2 y^4.
Using the chain rule on the left, and the chain and product rules on the right, we get
7 x^6 + 7y^6 dy dx = 14
x^2 d dx y
(^4) + y 4 d dx x
2
x^2 4 y^3 dy dx +^ y
(^4 2) x
Dividing both sides by 7 this becomes
x^6 + y^6 dy dx = 8x^2 y^3 dy dx
Now rearrange to get both terms with dydx on the same side:
x^6 − 4 xy^4 = 8x^2 y^3 dy dx − y^6 dy dx
8 x^2 y^3 − y^6
) (^) dy dx
Therefore dy dx = x
(^6) − 4 xy 4 8 x^2 y^3 − y^6 , or dy dx =^4 xy
(^4) − x 6 y^6 − 8 x^2 y^3 is also correct.
To get the slope of the curve at (7, 7) we plug in x = 7 and y = 7:
dy dx at (7,^ 7) equals^
3(i) Since x = t^2 − 1, x = 0 when t^2 = 1, so x = 0 when t = 1 and when t = −1. (Note that x does not approach 0 when t becomes infinite, unlike what happened in the project.)
3(ii) y = t^3 − t is also 0 when t = 1 and when t = −1, so the curve goes through the origin at those two values of t.
3(iii) We have dx dt = 2t, dy dt = 3t^2 − 1, and dy dx
dy dt dx dt
3 t^2 − 1 2 t
3(iv) When t = 1 the slope is 3 − 2 1 = 1, and when t = −1 the slope is (^3) −− 21 = −1. Therefore the curve has two slopes at the origin, 1 and −1.
Extra credit: y =
t^2 − 1
t = xt, so t = yx. Substituting this in the equation x = t^2 − 1 we get
x =
( (^) y x
y^2 x^2 −^1.
Multiplying through by x^2 this becomes
x^3 = y^2 − x^2 , or y^2 = x^2 + x^3.
- We have f ′(x) = x^3 (x − 12)^2 (x − 6) and f ′′(x) = 6x^2 (x − 12)(x − 9)(x − 4),
so sign charts for f ′(x) and f ′′(x) are
f ′(x) and f ′′(x)
From the first sign chart, f (x) is increasing for x < 0, for 6 < x < 12, and for x > 12, and f (x) is decreasing for 0 < x < 6. Since f (x) switches from increasing to decreasing at x = 0, x = 0 is a peak (local maximum of f (x)), and since f (x) switches from decreasing back to increasing at x = 6, x = 6 is a valley (local minimum of f (x)). From the second sign chart, f (x) is concave up for 4 < x < 9 and for x > 12, and f (x) is concave down for x < 0, for 0 < x < 4, and for 9 < x < 12. The sign changes of f ′′(x) are the inflection points of f (x), so these occur at x = 4, x = 9 and x = 12.
5(a) If a(x) = ln
sin x cos x
, then
a′(x) =
sin x cos x
d dx
sin x cos x =
cos x sin x
(cos x) (cos x) − (sin x) (− sin x) cos^2 x
= cos x sin x
cos^2 x + sin^2 x cos^2 x
cos x sin x
cos^2 x
sin x cos x
There are a couple of other ways to do this, depending on what you know. If you know that the derivative of tan x is sec^2 x, then you could write
a′(x) = d dx ln (tan x) =
tan x sec^2 x = sec^2 x tan x
