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For each of the following polynomials, determine if the given polynomial must have a global minimum, global maximum, or neither. Explain your reasoning.

a) f(x)= x²+3x+5
b) g(x)= -2x³+6x²-4x+1
c) h(x)= 4x⁴-8x³+2x²-6x+9
d) k(x)= -x²+4x-2

A) Polynomial must have a global minimum.
B) Polynomial must have a global maximum.
C) Polynomial has both global minimum and maximum.
D) Polynomial has neither global minimum nor maximum.

1 Answer

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Final answer:

To determine the global minimum or maximum, we analyze the degree and the leading coefficient of the polynomial. Quadratic polynomials have either a global minimum or maximum depending on the sign of the leading coefficient. Polynomials with odd degrees and a non-zero leading coefficient extend infinitely in opposite directions, having neither a global minimum nor maximum.

Step-by-step explanation:

When determining if a given polynomial must have a global minimum, global maximum, or neither, we assess the degree and leading coefficient of the polynomial.

a) f(x) = x² + 3x + 5: This is a quadratic polynomial with a positive leading coefficient. Such polynomials open upwards and therefore always have a global minimum but no global maximum.

b) g(x) = -2x³ + 6x² - 4x + 1: The leading term has a negative coefficient and an odd degree, indicating that the ends of the graph extend indefinitely in opposite directions. Hence, this polynomial has neither a global minimum nor maximum.

c) h(x) = 4x⁴ - 8x³ + 2x² - 6x + 9: This is a quartic polynomial with a positive leading coefficient. Since all even degree polynomials with positive leading coefficients tend towards positive infinity as x approaches both positive and negative infinity, this polynomial must have a global minimum but no global maximum.

d) k(x) = -x² + 4x - 2: As with the first example, this is a quadratic polynomial; however, the leading coefficient is negative, indicating the parabola opens downwards. Therefore, it has a global maximum and no global minimum.

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