Answer:
23) No
24) No
25) Yes
Step-by-step explanation:
Question 23)
We want to determine if a zero exists between 1 and 2 for the function:
[tex]f(x)=x^2-4x-5[/tex]
Find the zeros of the function. We can factor:
[tex]\displaystyle 0 = (x-5)(x+1)[/tex]
Zero Product Property:
[tex]x-5=0\text{ or } x+1=0[/tex]
Solve for each case. Hence:
[tex]\displaystyle x = 5\text{ or } x=-1[/tex]
Therefore, our zeros are at x = 5 and x = -1.
In conclusion, a zero does not exist between 1 and 2.
Question 24)
We have the function:
[tex]f(x)=2x^2-7x+3[/tex]
And we want to determine if a zero exists between 1 and 2.
Factor. We want to find two numbers that multiply to (2)(3) = 6 and that add to -7.
-6 and -1 suffice. Hence:
[tex]\displaystyle \begin{aligned} 0 & = 2x^2-7x + 3 \\ & = 2x^2 -6x -x + 3 \\ &= 2x(x-3) - (x-3) \\ &= (2x-1)(x-3) \end{aligned}[/tex]
By the Zero Product Property:
[tex]2x-1=0\text{ or } x-3=0[/tex]
Solve for each case:
[tex]\displaystyle x=\frac{1}{2} \text{ or } x=3[/tex]
Therefore, our zeros are at x = 1/2 and x = 3.
In conclusion, a zero does not exist between 1 and 2.
Question 25)
We have the function:
[tex]f(x)=3x^2-2x-5[/tex]
And we want to determine if a zero exists between -2 and 3.
Factor. Again, we want to find two numbers that multiply to 3(-5) = -15 and that add to -2.
-5 and 3 works perfectly. Hence:
[tex]\displaystyle \begin{aligned} 0&= 3x^2 -2x -5 \\ &= 3x^2 +3x - 5x -5 \\ &= 3x(x+1)-5(x+1) \\ &= (3x-5)(x+1)\end{aligned}[/tex]
By the Zero Product Property:
[tex]\displaystyle 3x-5=0\text{ or } x+1=0[/tex]
Solve for each case:
[tex]\displaystyle x = \frac{5}{3}\text{ or } x=-1[/tex]
In conclusion, there indeed exists a zero between -2 and 3.
