Questions: QUESTION 28 Consider the following function h: R → R, given by h(x)=2x^2+1 Determine which of the following statements about h is/are correct, if any. h is a function that is neither one-to-one nor onto. h is a bijective function. h is an onto function, but it is not one-to-one. h is a one-to-one function, but it is not onto.

QUESTION 28

Consider the following function h: R → R, given by
h(x)=2x^2+1

Determine which of the following statements about h is/are correct, if any.
h is a function that is neither one-to-one nor onto.
h is a bijective function.
h is an onto function, but it is not one-to-one.
h is a one-to-one function, but it is not onto.

Transcript text: QUESTION 28 Consider the following function $\mathrm{h}: \mathbb{R} \rightarrow \mathbb{R}$, given by \[ h(x)=2 x^{2}+1 \] Determine which of the following statements about h is/are correct, if any. $h$ is a function that is neither one-to-one nor onto. $h$ is a bijective function. h is an onto function, but it is not one-to-one. $h$ is a one-to-one function, but it is not onto.

Solution

Solution Steps

Step 1: Analyze the Function

The given function is $ h(x) = 2x^2 + 1 $. This is a quadratic function, which is a type of polynomial function. Quadratic functions are generally not one-to-one because they are symmetric about their vertex.

Step 2: Determine if the Function is One-to-One

A function is one-to-one if each output is produced by exactly one input. For quadratic functions like $ h(x) = 2x^2 + 1 $, the graph is a parabola opening upwards. This means that for any $ y $ value greater than 1, there are two corresponding $ x $ values (except at the vertex). Therefore, $ h(x) $ is not one-to-one.

Step 3: Determine if the Function is Onto

A function is onto if every possible output in the codomain is mapped to by some input in the domain. The range of $ h(x) = 2x^2 + 1 $ is $[1, \infty)$, since the minimum value of $ h(x) $ is 1 when $ x = 0 $. The codomain is $\mathbb{R}$, but the range does not cover all real numbers (it only covers numbers from 1 to infinity). Therefore, $ h(x) $ is not onto.