# Local minimin and maxmma

Assume that the polynomial P(x) has exactly two local maxima and one local minimum, and that these are the only critical points of P(x). Sketch possible graphs of P(x) and use them to answer the following.

(a) What is the largest number of zeros P(x) could have?

(b) What is the least number of zeros P(x) could have?

(c) What is the least number of inflection points P(x) could have?

(d) What is the smallest degree P(x) could have?

(e) What is the sign of the leading coefficient of P(x) ? (negative / positive)

### 1 Answer

- JacobLv 61 decade agoFavorite Answer
The shape of P(x) is just like the letter "M" with smooth shape.

(you can think of it by looking at the logo of McDonald.)

(a) P(x) could have at most 4 zeros.

i.e. when this "M" across the x-axis, just like -∩-∩-

(b) P(x) could have no zeros.

i.e. when the whole "M" is below the x-axis.

(c) P(x) could have at least 2 points of inflection.

i.e. In between those maximum and minimum.

Since at those max point, P"(x)<0, at those min point, P"(x)>0.

Therefore P"(x) has changed sign twice, and thus at least 2 zeros.

(d) P(x) is at least degree 4.

Since P(x) has 3 extrema, thus P'(x) has 3 zeros,

and so P'(x) is of at least degree 3, hence P(x) is of at least degree 4.

(e) The sign of the leading coefficient is negative.

Note that if P(x) is of "M" shape, then it must have even degree.

And the term of highest power will dominate the sign of P(x)

when x is going to positive infinity and negative infinity.

Now when x is going to either +ve or -ve infinity, P(x) is negative,

therefore the sign of leading coefficient must be negative.

Source(s): ME- Login to reply the answers