# MATHEMATICS 120 PROBLEM SET 3 - SOLUTIONS 1. The function f − ax

```MATHEMATICS 120 PROBLEM SET 3 - SOLUTIONS
1. The function f (x) is defined by

0 if − 2 ≤ x ≤ −1,



 ax2 + bx if − 1 &lt; x &lt; 2,
f (x) =

c if x = 2,



2x if 2 &lt; x ≤ 5.
Find all values a, b, c such that f (x) is continuous on [−2, 5].
For f (x) to be continuous at −1, we need 0 = a − b, so that a = b. For f (x) to be
continuous at 2, we need 4a + 2b = c = 4. Therefore c = 4, 4a + 2b = 6a = 4, a = 2/3,
b = 2/3.
2. Use the intermediate value theorem to prove that the equation x3 − 7x + 2 = 0 has three
solutions in the interval [−3, 3].
Let f (x) = x3 − 7x + 2, then f (−3) = −4 &lt; 0, f (−2) = 8 &gt; 0, f (2) = −4 &lt; 0,
f (3) = 8 &gt; 0. Since f (x) is continuous everywhere, by the intermediate value theorem
the equation f (x) = 0 has at least one solution in each of the intervals (−3, −2), (−2, 2),
(2, 3). (Note: since f (x) is a cubic polynomial, there cannot be more than 3 solutions.)
3. Give an example of a function f (x), defined on an open interval, such that f (x) has a
maximum value but does not have a minimum value. Also give an example of a function
g(x), defined for all x, such that g(x) has both a maximum value and a minimum value.
(You can give a formula or draw a graph – either is acceptable.)
If f (x) = −x2 on (−1, 1), then f (x) has a maximum value f (0) = 0 but no minimum
value. For g(x), we can take e.g. g(x) = sin x, with maximum value 1 and minimum value
−1. (The maximum and minimum values may be attained more than once – in fact if
g(x) = sin x, each of them is attained infinitely many times.)
4. Assume that the following is true for some function f (x), defined for all x: for every
&gt; 0 there is a δ &gt; 0 such that |f (x) − L| &lt; for some x such that 0 &lt; |x − a| &lt; δ (read
this carefully!). Does this imply that limx→a f (x) = L? Explain why or why not.
It does not. In the definition of the limit at a we have “for all x such that 0 &lt; |x − a| &lt; δ”,
not “for some x”. For example, let f (x) = sin(1/x) if x 6= 0 and f (0) = 0. Then f (x)
does not have a limit at 0. But if a = 0 and L = 0, then the statement in the problem
is true: for every &gt; 0 there is a δ &gt; 0 such that |f (x) − L| &lt; for some x such that
0 &lt; |x − a| &lt; δ, for example for x = 1/nπ with n large enough.
1
```