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sqrt(x)

Is it a "square root function"? Or exponential, since sqrt(x) = x^(1/2)? Or would it be more proper to call it an "inverse quadratic" since y=sqrt(x) is the inverse or y=x^2

2007-09-06 05:51:25 · 2 answers · asked by de4th 4 in Science & Mathematics Mathematics

2 answers

I think it is more properly called square root function.

Exponential functions are defined as a^x where the exponent is the variable and the base is a constant.

and it is not really an inverse quadratic either since y=x^2 does not really have an inverse (it is not a one-to-one function) if you're really trying to make an inverse out of y=x^2, then it would be y= +- sqrt(x).

2007-09-06 06:03:46 · answer #1 · answered by Derek C 3 · 0 0

It is the square root function. The square and square root functions are an inverse pair because y= x^2 and x=sqrt(y)

y = sinx and x= arcsin y is another inverse pair as is y=a^x and x = log base a (y). y= a^x is an exponential function and its inverse is a logarithmic function. You wouldn't call y=a^x an inverse logarithmic function would you?

2007-09-06 13:08:23 · answer #2 · answered by ironduke8159 7 · 0 0

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