Intermediate Value Theorem

Question

Using the intermediate value​ theorem, determine, if​ possible, whether the function f has at least one real zero between a and b.

f(x)=x^3+3x^2−9x−26​;

a=−7​,

b=-3



Answer 1

f^-1'(x) = 3x^2 + 6x -9 = 0

x^2 +2x -3 = 0

(x+3)(x-1) = 0

x = 1, -3 = turning points of the cubic equation, the positive zero is greater than 1, between 1 and 2

the negative zeros are between -1 and -2 and the other negative zero less than -3. the zero less than -3 is between -3 and -4 but much closer to -3.

by DeCartes' rule there is a maximum potential one real positive zero

and maximum 2 potential negative real zeros for this cubic there are exactly 1 positive real zero and 2 negative real zeros.

any 3 degree polynomial has at least 1 zero and potentially 3 zeros. Either 3 real zeros, or 1 real zero and 2 imaginary zeros. This polynomial has 3 real zeros, 1 positive, two negative.

Intermediate Value Theorem? IVT? whoops, forgot that. But results are the same.

Intermediate Value Theorem states that for an interval (a,b) there exists c such that a<c<b where f(a) <f(c) <f(b) if f is continuous. and true enough the cubic must intersect the x axis at least once, and maybe twice in the interval (-7, -3)

for this specific problem there are two zeros between -7 and -3