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The following values represent linear function ƒ(x) and rational function g(x). ƒ(3) = 7 g(3) = 5.6 ƒ(4) = 5 g(4) = 6.7

A. Determine whether or not there is a solution to the equation In 2-3 sentences describe whether there is a solution to the equation ƒ(x)=g(x) between x=3 and x=4.

B. Use complete sentences to justify your claim.

User Lenita
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2 Answers

6 votes
Pls find attached screen shot in part1 and 2
The following values represent linear function ƒ(x) and rational function g(x). ƒ(3) = 7 g-example-1
The following values represent linear function ƒ(x) and rational function g(x). ƒ(3) = 7 g-example-2
User Sprksh
by
7.1k points
4 votes

Answer with Explanation:

f(x)= Linear Function

g(x)= Rational Function

⇒Let , f(x)=a x +b

f(3)= 7

3 a + b=7------------(1)

f(4)=5

5 a + b=5--------(2)

⇒Equation(2) - Equation (1)

2 a= -2

Dividing both side by 2, we get

a = -1

Putting the value of 'a' in equation (1)

3 × (-1)+ b=7

b=7+3

b=10

So,Linear function = -x +10

⇒Let, Rational Function


g(x)=(1)/(px+q)\\\\g(3)=(1)/(3p+q)=5.6\\\\3.\rightarrow 3 p +q=(1)/(5.6)\\\\g(4)=6.7\\\\4.\rightarrow 4 p+q=(1)/(6.7)

Equation 4 - Equation 3


p=(1)/(6.7)-(1)/(5.6)\\\\ p=(-1.1)/(37.52)\\\\p=(-11)/(375.2)

Substituting the value of 'p' in equation 4


q=(10)/(6.7 * 5.6)=(100)/(375.2)

So, Rational Function


=(375.2)/(-11 x+100)

⇒Now, we have to check whether there is a solution to the equation

f(x)=g(x), between x=3 and x=4.


(375.2)/(-11 x+100)=-x+10\\\\\rightarrow 11 x^2-110 x-100 x+1000=375.2\\\\\rightarrow 11 x^2-210 x+624.80=0\\\\x=(210 \pm√(44100 -44 * 624.80))/(22)\\\\ax^2+bx+c=0\\\\x=\frac{-b\pm\Sqrt {b^2-4 ac}}{2a}\\\\x=(210\pm√(44100- 27491.20))/(22)\\\\x=(210 \pm 128.875)/(22)\\\\x=(338.875)/(22) \\\\x=(81.125)/(22)\\\\x=15.40 \text{and} \\\\x=3.6875

Yes, there is a solution to the equation ƒ(x)=g(x) between x=3 and x=4 which is, x=3.6875.

User TimW
by
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