Question

In Exercises 5-12, (a) identify the domain and range and (b) sketch the graph of the function. $$y=1+\frac{1}{x^{2}}$$

Short answer

The domain of the function \(y=1+\frac{1}{x^{2}}\) is all real numbers except zero, that is, \((-\infty,0) \cup (0, +\infty)\) or \(x \neq 0\). The range is \((1,+\infty)\). The graph of the function is a reciprocal graph shifted upwards by one unit which approaches value of 1 as \(x\) tends to either from \(+\infty\) or \(-\infty\), and excludes the point where \(x=0\), causing an asymptote.

Step-by-step solution

Identify the domain

The domain of a function are all the values that x can take. In this case, as \(x^{2}\) is in the denominator, the denominator becomes zero when \(x=0\). So, this function is defined for all values of x, except \(x=0\). Thus, the domain of the function is \((-\infty,0) \cup (0,+\infty)\) in interval notation or \(x \neq 0\) in inequality notation.

Identify the range

The range of a function are all the possible values of y. The equation is \(y=1+\frac{1}{x^{2}}\), where the second term is always non-negative as squares are always positive and we're adding 1. Hence the minimum value of y is when \(x\) tends to infinity, we get \(y=1\). And as y increases as x approaches to 0 from left or right side, therefore, the range of the function is \((1,+\infty)\).

Sketch the graph of the function

The graph of the function \(y=1+\frac{1}{x^{2}}\) is a reciprocal graph shifted one unit upwards. To sketch it, lay out a grid for values of \(x\) and \(y\), remembering to exclude the point where \(x=0\), this causes an asymptote. Then plot some values for positive and negative x. The sketch should show two 'branches' of the graph approaching the value of 1 as x diverges.