Suppose f(x) = x^{6}+3x+1 . In this problem, we will show that f has
exactly one root (or zero) in the interval \lbrack -5, -1 \rbrack .

(a) First, we show thatf has a root in the interval (-5, -1) .
Since f is a
choose
continuous
differentiable
polynomial
function on the
interval \lbrack -5, -1 \rbrack and
f(-5) = and
f(-1) = ,
the graph of y = f(x) must cross the x -axis
at some point in the interval (-5, -1) by the
choose
intermediate value theorem
mean value theorem
squeeze theorem
Rolle's theorem
.
Thus, f has at least one root in the interval
\lbrack -5, -1 \rbrack .

(b) Second, we show thatf cannot have more than one
root in the interval \lbrack -5, -1 \rbrack by a thought experiment.
Suppose that there were two roots x = a and
x = b in the interval \lbrack -5, -1 \rbrack with a < b . Then
f(a) = f(b) = .
Since f is
choose
continuous
differentiable
polynomial
on the interval \lbrack -5, -1 \rbrack and
choose
continuous
differentiable
polynomial
on the interval (-5, -1) ,
by
choose
intermediate value theorem
mean value theorem
squeeze theorem
Rolle's theorem
there would exist a point c in interval (a,b)
so that f'(c) = 0 .
However, the only solution to f'(x) = 0 is
x = , which is not
in the interval (a,b) , since (a,b) \subseteq \lbrack -5, -1 \rbrack .
Thus, f cannot have more than one root in \lbrack -5, -1 \rbrack .

(a) First, we show that

(b) Second, we show that

(Note: where the problem asks you to make a choice select the ** weakest ** choice that works in the given context. For example "continuous" is a weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the mean value theorem because Rolle's theorem applies to fewer cases.)

You can earn partial credit on this problem.