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Chris and Jen decide to go apple picking at a local apple orchard. As the number of apples they pick increases the cost of the apples also increases. The function ff relates the varying cost of Chris and Jen's apples, cc , in terms of the varying number of pounds of apples that Chris and Jen pick, nn , where c=f(n)c=f(n) and ff is defined by f(n)=0.4n+6f(n)=0.4n+6.
Evaluate f−1(33) Note( this is F^-1)
Determine the rule for the function f^-1


Sagot :

Any relation that has an inverse is a function

  • The value of [tex]\mathbf{f^{-1}(33)}[/tex] is [tex]\mathbf{f^{-1}(33) = 67.5}[/tex]
  • The rule of [tex]\mathbf{f^{-1}(n)}[/tex] is [tex]\mathbf{f^{-1}(n) = 2.5(n -6)}[/tex]

The function is given as:

[tex]\mathbf{f(n) = 0.4n + 6}[/tex]

(a) Evaluate [tex]\mathbf{f^{-1}(33)}[/tex]

First, we calculate the inverse function

We have:

[tex]\mathbf{f(n) = 0.4n + 6}[/tex]

Rewrite as:

[tex]\mathbf{y = 0.4n + 6}[/tex]

Subtract 6 from both sides

[tex]\mathbf{y -6= 0.4n + 6 - 6}[/tex]

[tex]\mathbf{y -6= 0.4n}[/tex]

Divide both sides by 0.4

[tex]\mathbf{\frac{1}{0.4}(y -6)= \frac{0.4n}{0.4}}[/tex]

[tex]\mathbf{\frac{1}{0.4}(y -6)= n}[/tex]

[tex]\mathbf{2.5(y -6)= n}[/tex]

Make n the subject

[tex]\mathbf{n = 2.5(y -6)}[/tex]

Rewrite as:

[tex]\mathbf{n = 2.5(f(n) -6)}[/tex]

So, the inverse function is:

[tex]\mathbf{f^{-1}(n) = 2.5(n -6)}[/tex]

Substitute 33 for n to calculate [tex]\mathbf{f^{-1}(33)}[/tex]

[tex]\mathbf{f^{-1}(33) = 2.5(33 -6)}[/tex]

[tex]\mathbf{f^{-1}(33) = 2.5(27)}[/tex]

[tex]\mathbf{f^{-1}(33) = 67.5}[/tex]

(b) The rule of [tex]\mathbf{f^{-1}(n)}[/tex]

In (a), we have: [tex]\mathbf{f^{-1}(n) = 2.5(n -6)}[/tex]

Hence, the rule of [tex]\mathbf{f^{-1}(n)}[/tex] is [tex]\mathbf{f^{-1}(n) = 2.5(n -6)}[/tex]

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