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anonymous

  • 5 years ago

find the inverse function of f state the domain and range of f f(x) = 3x-2/x+5

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  1. anonymous
    • 5 years ago
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    Is your function,\[f(x)=\frac{3x-2}{x+5}\]?

  2. anonymous
    • 5 years ago
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    By the definition of inverse function, if g(x) is the inverse of f(x), then\[f(g(x))=x=g(f(x))\]To find your inverse, you set\[f(g(x))=\frac{3g(x)-2}{g(x)+5}:=x\]and solve for g(x). Doing this you obtain\[g(x)=\frac{5x+2}{1-x}\]It is the case that the domain of a function is equal to the range of the inverse function, and the range of the original function is equal to the domain of the inverse.

  3. anonymous
    • 5 years ago
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    The domain of f is the set of all x for which f(x) is defined. In this case, f is defined for all real x except where x=-5 (the function is not defined there since the denominator will be zero). So,\[D_f=\left\{ x \in \mathbb{R} | x \ne -5 \right\}\]

  4. anonymous
    • 5 years ago
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    The range of f is equal to the domain of g. Since g is defined for all real x except x=1 (since then, the denominator is 0), you have\[D_g=\left\{ x \in \mathbb{R} | x \ne 1 \right\}=R_f\]i.e. this is the range of f.

  5. anonymous
    • 5 years ago
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    And I think I just worked out the wrong inverse...

  6. anonymous
    • 5 years ago
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    i dont see 5x+2/1-x options are a. 5x+2/3+x b. 5x+2/3-x c. 3x+2/x-5 d. x+5/3x-2

  7. anonymous
    • 5 years ago
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    The inverse function should have been\[g(x)=\frac{5x+2}{3-x}\]

  8. anonymous
    • 5 years ago
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    b)

  9. anonymous
    • 5 years ago
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    Domain of f is the same as I've written above. Range of f is equal to domain of inverse, and domain of inverse is equal to all x real except x=3, so the range of f is\[R_f=\left\{ x \in \mathbb{R}|x \ne 3 \right\}\]

  10. anonymous
    • 5 years ago
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    thanks a lot man

  11. anonymous
    • 5 years ago
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    could use a new fan:D

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