Question Video: Identifying Functions without an Inverse over Their Entire Domain | Nagwa Question Video: Identifying Functions without an Inverse over Their Entire Domain | Nagwa

Question Video: Identifying Functions without an Inverse over Their Entire Domain Mathematics • Second Year of Secondary School

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Which of the following functions does not have an inverse over its whole domain? Assume that the codomain of each function is equal to its range. [A] π(π₯) = 2π₯ [B] π(π₯) = 2^(π₯) [C] π(π₯) = 1/π₯ [D] π(π₯) = π₯^(2)

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Video Transcript

Which of the following functions does not have an inverse over its whole domain? Assume that the codomain of each function is equal to its range. Is it option (A) π of π₯ is equal to two π₯? Option (B) π of π₯ is equal to two to the power of π₯. Option (C) π of π₯ is equal to one over π₯. Or option (D) π of π₯ is equal to π₯ squared.

To answer this question, we need to consider the conditions under which a function has an inverse. A function only has an inverse over its entire domain if it is a one-to-one function. This means that every input to the function produces a unique output so that when we go the other way using the inverse function, every input to this function also produces a unique output. We can work out which of our four functions isnβt one-to-one by sketching graphs of each of them.

For option (A), the graph of π¦ equals π of π₯ or π¦ equals two π₯ is a straight-line graph through the origin. This is clearly a one-to-one function over its entire domain. Every input has a unique output, and so going the other way, the same will be true. More generally though, we can determine whether a function is one to one by drawing horizontal and vertical lines on its graph. If each horizontal and each vertical line intersects the graph a maximum of one time, then the function is one to one.

Letβs now consider option (B). The graph of π¦ equals two to the power of π₯ is an exponential graph passing through the point zero, one with a horizontal asymptote along the π₯-axis. This is also a one-to-one function over its entire domain because every vertical line intersects the graph once and every horizontal line intersects the graph a maximum of one time. We have therefore ruled out options (A) and (B). Now, letβs consider option (C).

For the graph of π¦ equals one over π₯, there are two branches, one in the first quadrant, where π₯ and π¦ are both positive, and one in the third quadrant, where π₯ and π¦ are both negative. The π₯- and π¦-axes are horizontal and vertical asymptotes. And we can once again see that by drawing horizontal and vertical lines, the function is one to one over its entire domain.

This leaves us with just option (D). The graph of π¦ equals π₯ squared is a positive parabola with its minimum point at the origin. We can see that if we draw horizontal lines across the graph, some of these lines will intersect the graph in more than one place. This means that for certain output values, there are two possible input values. And so the function π of π₯ isnβt one to one. This means we canβt find the inverse of π of π₯ over its entire domain because we donβt know which of those input values the output value should be mapped to by the inverse function.

The correct answer is therefore option (D). The function that does not have an inverse over its whole domain is π of π₯ is equal to π₯ squared.

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