Q&A for How to Find the Inverse of a Quadratic Function

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Find the inverse of (1 - 2x)^3

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First, let me point out that this question is beyond the scope of this particular article. The article is about quadratic equations, which implies that the highest exponent is 2. Your question presents a cubic equation (exponent =3). Nevertheless, basic algebra allows you to find the inverse of this particular type of equation, because it is already in the "perfect cube" form. First, set the expression you have given equal to y, so the equation is y=(1-2x)^3. Then invert it by switching x and y, to give x=(1-2y)^3. Now perform a series of inverse algebraic steps to solve for y. These steps are: (1) take the cube root of both sides to get cbrt(x)=1-2y [NOTE: I am making up the notation “cbrt(x) to mean “cube root of x” since I can’t show it any other way here]; (2) Subtract 1 from both sides to get cbrt(x)-1=-2y; (3) Divide both sides by -2 to get (cbrt(x)-1)/-2=y; (4) simplify the negative sign on the left to get (1-cbrt(x))/2=y. The final equation should be (1-cbrt(x))/2=y. This is your inverse function.

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Where can I find more examples so that I know how to set up and solve my homework problems?

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The Internet is filled with examples of problems of this nature. Google "find the inverse of a quadratic function" to find them.

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How do I find the inverse of f(x)=1/(sqrt(x^2-1)?

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Begin by switching the x and y terms (let f(x)=y), to get x=1/(sqrt(y^2-1). Then perform basic algebraic steps to each side to isolate y. Without getting too lengthy here, the steps are (1) square both sides to get x^2=1/(y^2-1); (2) transpose numerators and denominators to get y^2-1=1/x^2; (3) add 1 to both sides to get y^2=(1/x^2)+1; (4) square root both sides to get y=sqrt((1/x^2)+1).

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How do I state and give a reason for whether there's an inverse of a function?

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If the function is one-one & onto/bijective, it has an inverse. So: ONE ONE/SURJECTIVE:let a,b belong to the given domain such that f(a)=f(b). Then, if after working it out, a=b, the function is one one/surjective. To check whether it's onto, let y=f(x) and solve to see whether all values of y lie in the range of the fn.

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