Midterm 1 Solutions. March 5, 2009. 1. [2 pts each] Find the derivative of each of
the following functions. (a) f(x) = cot x. Solution: Write cot x = cos x sin x and use ...
Midterm 1 Solutions March 5, 2009 1. [2 pts each] Find the derivative of each of the following functions. (a) f (x) = cot x Solution: Write cot x =
cos x sin x
and use the quotient rule:
d cot x = dx
d dx
d cos x sin x − cos x dx sin x 2
sin x =−
=
− sin2 x − cos2 x sin2 x
1 = − csc2 x. sin2 x
(b) f (x) =
sin x x
Solution: Again use the quotient rule: d sin x = dx x (c)
d dx
d sin x x − sin x dx x x cos x − sin x = . 2 x x2
√ √ f (x) = ( x − x − 1)( x + x + 1) Solution: First multiply everything out using (a − b)(a + b) = a2 − b2 and a =
√
x, b = x + 1.
We get f (x) = x − (x + 1)2 = x − x2 − 2x − 1 = −x2 − x − 1. Therefore f 0 (x) = −2x − 1. 2. [2 pts each] Let f (x) =
√
x − 1.
(a) Find the derivative of f using the limit-definition of the derivative. Solution: √ √ f (x + h) − f (x) x+h−1− x−1 = lim h→0 h→0 h h √ √ √ √ x+h−1− x−1 x+h−1+ x−1 √ √ = lim h→0 h x+h−1+ x−1 f 0 (x) = lim
= lim
h→0
h x+h−1−x+1 √ √ √ = lim √ h( x + h − 1 + x − 1) h→0 h( x + h − 1 + x − 1) = lim √ h→0
1 1 √ = √ . 2 x−1 x+h−1+ x−1
(b) What is the domain of the derivative. Why? Solution: The derivative function has two problems: there is a root and there is a denominator. The term under the root cannot be negative. This means x−1≥0 x ≥ 1. However, when x = 1, the denominator is zero (this is the second problem). Therefore x 6= 1. This gives that the derivative is defined for all x > 1. This is the domain of the function f 0 (x). (c) Find the equation of the line tangent to the graph of f at x = 2. Solution: We need two pieces of information: the slope of the tangent line at x = 2 and the point on the graph at x = 2. The first of these is given by the derivative. The slope of the tangent line is 1 f 0 (2) = √ = 1/2. 2 2−1 The point on the graph is given by (2, f (2)) with f (2) =
√
2 − 1 = 1.
Therefore the equation of the tangent line is y − 1 = (1/2)(x − 2) y = x/2. 3. [3 pts each] Find the following limits:
(a) lim
x→0
sin(2x) tan(3x)
Solution: sin(2x) sin(2x) cos(3x) = lim = lim x→0 tan(3x) x→0 x→0 sin(3x) lim
2x cos(3x) x→0 3x
= lim
sin(2x) 2x sin(3x) 3x
2x 2x
sin(2x) cos(3x) 3x 3x sin(3x) sin(2x)
=
2 2x lim cos(3x) sin(3x) 3 x→0 3x
Since sin(3x) sin(2x) = 1, lim = 1, and lim cos(3x) = 1, x→0 x→0 2x 3x we get that the limit is 2/3. lim
x→0
(b) lim
x→−2
1 x
+ 21 x+2
Solution:
lim
x→−2
2+x + 12 2+x 1 = lim 2x = lim = lim = −1/4. x→−2 x + 2 x→−2 (2x)(x + 2) x→−2 2x x+2
1 x
4. [3 pts each] Find all values of c so that the function f given by ( f=
cx + 1 cx2
x≥2 x 2. Therefore,
lim f (x) = lim+ cx + 1.
x→2+
x→2
But the function given by cx + 1 is continuous for all x. In particular it has right continuous at each x. This gives lim cx + 1 = 2c + 1.
x→2+
This was the value of f (2). This verifies that f is right continuous no matter which value of c we use. Answer: all values of c. (b) is left continuous at x = 2 (but not necessarily right continuous). Solution: In order that f is left continuous at x = 2, it must be that lim f (x) = f (2).
x→2−
Again, since f (2) = 2c + 1, we would like to have lim f (x) = 2c + 1.
x→2−
By reasoning similar to that used in part (a), we see that lim f (x) = 4c.
x→2−
Therefore we would like 2c + 1 = 4c c = 1/2. 5. [2 pts each] (a) Draw a sketch of a function f which has the following properties at x = 0: right continuous, not left continuous, has both a left and right limit but has no limit. Solution:
(b) Draw a sketch of a function f which has the following properties at x = 0: right continuous, left continuous, not differentiable. Solution:
(c) Draw a sketch of a function f which has the following properties at x = 0: not right continuous, not left continuous, has a left limit but no right limit. Solution:
