a.
To Solve:
The inequality f(x)>0 .
x∈ℝ
Given:
f(x)=(x2+4)(2x2+3)
Concepts Used:
The zeroes of a polynomial function are the set of zeroes of all its factors.
Polynomials are continuous everywhere in ℝ . So they can change signs only at the points where they cross the x -axis.
The solution to f(x)>0is the region where the curve y=f(x) lies above the x -axis. This represents the region where f(x) is positive.
Calculations:
Determine the zeroes of the factor x2+4 of the polynomial function f(x)=(x2+4)(2x2+3) .
x2+4=0⇒x2=−4
Note that x2+4 cannot be zero for any real number x because there is no real number whose square is a negative number.
Determine the zeroes of the factor 2x2+3 of the polynomial function f(x)=(x2+4)(2x2+3) .
2x2+3=0⇒2x2=−3⇒x2=−32
Note that 2x2+3 cannot be zero for any real number x because there is no real number whose square is a negative number.
The polynomial function f(x)=(x2+4)(2x2+3) has no zero in ℝ . Thus, it can be said that f(x) cannot change its sign.
Determine the sign of f(x)=(x2+4)(2x2+3) in ℝ . Take a test point x=0∈ℝ and evaluate f(x) :
f(x)=(x2+4)(2x2+3)⇒f(0)=(0+4)(0+3)⇒f(0)=(4)(3)⇒f(0)=12⇒f(0)>0
f(x) is positive in ℝ .
Conclusion:
The set x∈ℝis the solution for the inequality f(x)>0 .
b.
To Solve:
The inequality f(x)≥0 .
x∈ℝ
Given:
f(x)=(x2+4)(2x2+3)
Concepts Used:
Polynomials are continuous everywhere in ℝ . So they can change signs only at the points where they cross the x -axis.
The solution to f(x)≥0is the region where the curve y=f(x) lies on or above the x -axis. This represents the region where f(x) is not negative.
For a polynomial function f(x) , the solution to f(x)>0 is a subset of the solution to f(x)≥0 . The solution to f(x)≥0 is the union of solution to f(x)>0 and solution set of f(x)=0 .
Known from previous part:
The polynomial function f(x)=(x2+4)(2x2+3) has no zero in ℝ .
f(x) is positive in ℝ . The solution to f(x)>0 is x∈ℝ .
Calculations:
f(x) is positive in ℝ so the solution set of f(x)=0 is null.
It follows that the solution to f(x)≥0 equals the solution to f(x)>0 , namely, ℝ .
Conclusion:
The solution of f(x)≥0 is x∈ℝ .
c.
To Solve:
The inequality f(x)<0 .
No solution.
Given:
f(x)=(x2+4)(2x2+3)
Concepts Used:
Polynomials are continuous everywhere in ℝ . So they can change signs only at the points where they cross the x -axis.
The solution to f(x)≥0is the region where the curve y=f(x) lies on or above the x -axis. This represents the region where f(x) is not negative.
The solution to f(x)>0 is a subset of the solution to f(x)≥0 . The solution to f(x)≥0 is the union of solution to f(x)>0 and solution set of f(x)=0 .
For a polynomial function f(x) , the solution to f(x)<0 is the complement of solution set f(x)≥0 considering ℝ as universal set.
Known from previous part:
The solution to f(x)≥0 is x∈ℝ .
Calculations:
The complement of ℝ is ℝ−ℝ=ϕ .
It follows that the solution to f(x)<0 is the empty set .
Conclusion:
There is no solution to f(x)<0 .
d.
To Solve:
The inequality f(x)≤0 .
No Solution.
Given:
f(x)=(x2+4)(2x2+3)
Concepts Used:
Polynomials are continuous everywhere in ℝ . So they can change signs only at the points where they cross the x -axis.
The solution to f(x)≥0is the region where the curve y=f(x) lies on or above the x -axis. This represents the region where f(x) is not negative.
The solution to f(x)>0 is a subset of the solution to f(x)≥0 . The solution to f(x)≥0 is the union of solution to f(x)>0 and solution set of f(x)=0 .
For a polynomial function f(x) , the solution to f(x)≤0 is the complement of solution set f(x)>0 considering ℝ as the universal set.
Known from previous part:
The solution to f(x)>0 is x∈ℝ .
Calculations:
The complement of ℝ is ℝ−ℝ=ϕ .
It follows that the solution to f(x)≤0 is the empty set .
Conclusion:
There is no solution to f(x)≤0 .
Chapter 2 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
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