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Unformatted text preview: Solutions to Quiz 5A
February 25, 2008
1. Find the equation of the tangent line to the curve f (x) =
point (0, −1). x3 − 2
x+2 at the First we need to ﬁnd f (x) to compute the slope of the tangent line.
Here we will use the quotient rule.
(x + 2) dx (x3 − 2) − (x3 − 2) dx (x + 2)
f (x) =
(x + 2)2
(x + 2) ∗ (3x2 ) − (x3 − 2) ∗ 1
(x + 2)2 Next, we need to ﬁnd the slope of the tangent line at (0, −1), so the
m = f (0) = (0 + 2) ∗ (02 ) − (03 − 2) ∗ 1
(0 + 2)
2 Recall that y = m ∗ x + b where b is the y -intercept. Since the function
goes through (0,-1) we already know that b = −1. Thus, the line is
y = 1/2x − 1.
NOTE: Notice that I did not waste time in reducing or rewriting f (x).
Once I found the found the derivative, I stopped and plugged in the
value that I needed. Only reduce or simplify when it is asked of you or
when it will make the problem easier. Second, make sure that you are
careful with your signs!
1 2. Find the derivative of the function:
(a) f (x) = x2 cos(x)
f (x) = x2
(b) g (x) = e(x d
cos(x) + cos(x) x2 = −x2 sin(x) + 2x cos(x)
dx 3) g (x) = ex 3 d3
x = 3x2 ex
dx (c) h(x) = sin(e)
h (x) = 0
Notice that sin(e) is just a number, so the derivative is zero. If
you do not believe me, type it in your calculator!
3. Evaluate the following limit: limx→0 sin(3x)
x sin(3x) 3
∗ = 3 ∗ lim
lim 2 ...
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