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Chapter 1: Problem 64

Find the domain of the function. $$h(x)=\frac{10}{x^{2}-2 x}$$

Short Answer

Expert verified
The domain of the given function is all real numbers except \(x=0\) and \(x=2\).

Step by step solution

01

Identify the Denominator

Our first step is to identify the denominator of the function, which is \(x^{2}-2 x\).
02

Set the Denominator Equal to Zero

We need to find which x-values make the denominator equal to zero. To do so, we set the denominator equal to zero. So, we solve the equation \(x^{2}-2 x=0\).
03

Solve for x

Factor x out of the equation to get \(x*(x-2)=0\). This equation equals zero if either of the factors equals zero. Therefore, we set \(x=0\) and \(x-2=0\) and solve for x in each case. The solutions are \(x=0\) and \(x=2\).
04

Exclude these values from the domain

Finally, the values obtained should be excluded from the domain of the function since these make the denominator zero. Hence, the domain of the function \(h(x)=\frac{10}{x^{2}-2 x}\) is all real numbers excluding \(x=0\) and \(x=2\).

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Most popular questions from this chapter

Use the functions \(f(x)=\frac{1}{8} x-3\) and \(g(x)=x^{3}\) to find the indicated value or function. $$\left(g^{-1} \circ f^{-1}\right)(-3)$$

Proof Prove that if \(f\) and \(g\) are one-to-one functions, then \((f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x)\).

Determine whether the statement is true or false. Justify your answer. It is possible for an odd function to have the interval \([0, \infty)\) as its domain.

Prove that the product of two odd functions is an even function, and that the product of two even functions is an even function.

Prove that a function of the following form is even. $$y=a_{2 n} x^{2 n}+a_{2 n-2} x^{2 n-2}+\cdots+a_{2} x^{2}+a_{0}$$
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