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Express f(x) in the form a + b/(x+1), where a and b are constants. Hence, give a sequence of three transformations which take the graph of y = 1/x onto the graph of y = f(x). State the range of f. Form the composite function f(f(x)). Hence, or otherwise, obtain an expression for f⁻¹(x).

Mathematics

ScanSolve AI Answer

Answer

f(x)=3−1x+1f(x) = 3 - \frac{1}{x+1}

Step-by-step solution

4 steps

To express f(x)=3x+2x+1f(x) = \frac{3x+2}{x+1} in the form a+bx+1a + \frac{b}{x+1}, we can use algebraic manipulation.

  1. Rewrite the numerator in terms of the denominator.

    We want to create a term of (x+1)(x+1) in the numerator.

    3x+2=3(x+1)−3+23x+2 = 3(x+1) - 3 + 2 3x+2=3(x+1)−13x+2 = 3(x+1) - 1
  2. Substitute this back into the expression for f(x)f(x).

    f(x)=3(x+1)−1x+1f(x) = \frac{3(x+1) - 1}{x+1}
  3. Separate the terms.

    f(x)=3(x+1)x+1−1x+1f(x) = \frac{3(x+1)}{x+1} - \frac{1}{x+1} f(x)=3−1x+1f(x) = 3 - \frac{1}{x+1}
  4. Compare with the given form a+bx+1a + \frac{b}{x+1}.

    By comparing f(x)=3−1x+1f(x) = 3 - \frac{1}{x+1} with f(x)=a+bx+1f(x) = a + \frac{b}{x+1}, we find the values of aa and bb.

    a=3a = 3 b=−1b = -1

    The expression for f(x)f(x) in the required form is:

    f(x)=3−1x+1\boxed{f(x) = 3 - \frac{1}{x+1}}

    where a=3a=3 and b=−1b=-1.

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