MATH 1500 Problem Workshop 8 Solutions
1. We need to show for any x2 > x1 on the interval (a, b), that x2 = x1 .
Let [x1 , x2 ] be an interval. We are told that f (x) = 0 on (a, b) we therefore the function
is dierentiable on (x1 , x2 ). Since f is dieren
MATH 1500 Problem Workshop 9 Solutions
1. (a) We can solve for the derivative to be f (x) = 2e2x ex = e2x (2 ex ) and
therefore f (x) = ex 4e2x = e2x (ex 4). We saw in Question 4a that the
critical number was ln 2. Hence on a number line we would put the
MATH 1500 Problem Workshop 7 Solutions
1. (a)
2x + 5
is of non zero/0 form and therefore goes to either innty or
x2
x+2
negative innity. The numerator has a limit of
lim
lim 2x + 5 = 9
x2
and the denominator is negative as x 2 . Therefore the limit goes
MATH 1500 Problem Workshop 10 Solutions
1. The domain is cfw_x : x = 0. The y-intercept doesnt exist since x = 0. The x-intercept is
when y = 0, so x2 + x + 1 = 0 which has no solution since b2 4ac = 1 4 = 3
in the quadratic formula. The horizontal asympt
MATH 1500 Problem Workshop 12 Solutions
1. The denition of area (when f (x) > 0) is
n
A = lim f (x1 ) + f (x2 ) + + f (xn ) x = lim
n
f (xi )x
n
where
x =
2
ba
=
n
n
and
i=1
xi = a + ix =
2i
.
n
Thus the area is
n
A = lim
n
f (xi )x
i=1
n
= lim
n
i=1
n
=