Question

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

Answer 1

Pls find attached screen shot in part1 and 2

Answer 2

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

     $g(x)=\frac{1}{px+q}\\\\g(3)=\frac{1}{3p+q}=5.6\\\\3.\rightarrow 3 p +q=\frac{1}{5.6}\\\\g(4)=6.7\\\\4.\rightarrow 4 p+q=\frac{1}{6.7}$

Equation 4 - Equation 3

 $p=\frac{1}{6.7}-\frac{1}{5.6}\\\\ p=\frac{-1.1}{37.52}\\\\p=\frac{-11}{375.2}$

Substituting the value of 'p' in equation 4

$q=\frac{10}{6.7 \times 5.6}=\frac{100}{375.2}$  

So, Rational Function

         $=\frac{375.2}{-11 x+100}$    

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

    f(x)=g(x), between x=3 and x=4.

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

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