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Chapter 5: Problem 86

The derivative of the function has the same sign for all \(x\) in its domain, but the function is not one-to-one. Explain. \(f(x)=\frac{x}{x^{2}-4}\)

Short Answer

Expert verified
The derivative of the function \(f(x)=\frac{x}{x^{2}-4}\) is \(f'(x)=\frac{4}{(x^2-4)^2}\), which is always positive within its domain (\(x > -2, x < 2, x > 2\)), indicating that the function is always increasing. However, the function is not one-to-one because as x approaches \(-2\) or \(2\), the same y-values are produced for different x-values.

Step by step solution

01

Find the derivative of the function

As the derivative represents the rate of change of a function, deriving \(f(x)\) will allow us to assess whether it's increasing or decreasing. The quotient rule can be used here, which is \(\frac{d}{dx}[\frac{v}{u}]=\frac{vu'-uv'}{u^2}\). Using this rule, the derivative \(f'(x)\) of the function can be found as follows: \[f'(x)=\frac{x^{2}(1)-1(x^{2}-4)}{(x^{2}-4)^{2}}=\frac{4}{(x^2-4)^2}\]
02

Analyze the sign of the derivative

The derivative \(f'(x)=\frac{4}{(x^2-4)^2}\) is positive within its domain, due to square in the numerator and the denominator. The function鈥檚 domain excludes the points \(x=-2\) and \(x=2\), because that鈥檚 where the denominator would equal zero, causing the function to be undefined. Hence, \(f'(x)>0\) for \(x\) in the domain of \(f(x)\). This confirms that the function should always be increasing within its domain.
03

Study the original function in relation to one-to-one

Even though the function is always increasing, it is not one-to-one. For a function to be classified as one-to-one, every y-value must correspond to exactly one x-value. In the case of our function \(f(x)=\frac{x}{x^{2}-4}\), as x approaches \(-2\) or \(2\), \(f(x)\) tends to \(\infty\) or \(-\infty\), respectively. This means the function will generate the same y-values for different x-values and will not be one-to-one.

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