Diagnostic Quiz for MATH 1272: Calculus II
This quiz does not affect your grade at all, and is only for you to assess whether
or not 1272 is right for you.
√
1. Let f (x) = 15 − 2x − x2 .
(a) Factor the polynomial 15 − 2x − x2 into linear factors.
15 − 2x − x2 = −(x2 + 2x − 15) = −(x + 5)(x − 3)
(b) Identify the largest possible domain D for f .
Because the value under the square root must be non-negative to return a real
number, we must have 15 − 2x − x2 ≥ 0. The previous problem shows that
15 − 2x − x2 = 0 when x = −5, 3, and it is easy to see that it is positive between
these values, and negative everywhere else. So D = [−5, 3].
(c) Compute f 0 (x).
1
Write f (x) = (15 − 2x − x2 ) 2 ; then by the chain rule,
1
1
−1 − x
f 0 (x) = (15 − 2x − x2 )− 2 (−2 − 2x) = √
.
2
15 − 2x − x2
(d) Find the equation for tangent line to the graph of y = f (x) at x = −2.
p
√
We compute: f (−2) = 15 − 2(−2) − (−2)2 = 15, and f (−2) = √115 . The
general formula for the tangent line is y − f (−2) = f 0 (−2)(x − (−2)), giving us
y−
√
1
15 = √ (x + 2)
15
(e) What is the maximum value of f on D? At what value of x does it occur?
Using the formula from part (c), we see that the only solution to f 0 (x) = 0
is
p when x = −1; this is either a local max or a local min. Since f (−1) =
15 − 2(−1) − (−1)2 = 4 is greater than the values of f (x) at −5 and 3 (that is,
f (x) = 0 there), this must be a local max. Since there is only one point where the
derivative is 0, and it is greater than the value of the function at the endpoints of
its domain, it is a global max: x = −1, f (−1) = 4.
1
(f) What is the minimum value of f on D? At what values of x does it occur?
From the previous part, we see that there are no minima x with f 0 (x) = 0;
therefore minima must occur at the boundary of the domain: x = −5, 3, where
f (x) = 0. Since square roots are always non-negative, these are certainly minimum values.
√
(g) Sketch a graph of y = f (x) = 15 − 2x − x2 .
It’s a half-circle of radius 4, centered at (−1, 0), lying above the x-axis.
(h) Can you compute
R3
−5
f (x)dx without computing the antiderviative of f ?
By the previous description, it is half the area of a circle of radius 4, or 12 π42 = 8π.
2. Let f be a continuous function defined on the interval [a, b], and define a new function
g on [a, b] by the formula
Z x
g(x) =
f (t)dt.
a
0
What is the derivative g (x) when x ∈ (a, b)? Why?
The fundamental theorem of calculus tells us that g 0 (x) = f (x).
3. Compute the indefinite integral
Z
dx
.
5 − 3x
We’ll use u-substitution, with u = 5 − 3x. Then du = −3dx, so
Z
Z
dx
1
du
1
1
=−
= − ln |u| + C = − ln |5 − 3x| + C
5 − 3x
3
u
3
3
2