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Found in: Page 120

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# What is the image of a function f from ${\mathbf{ℝ}}$ to ${\mathbf{ℝ}}$ given by ${\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbit{t}}}^{{\mathbf{3}}}{\mathbf{+}}{\mathbit{a}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbit{b}}{\mathbit{t}}{\mathbf{+}}{\mathbit{c}}$ ,where a,b,c are arbitrary scalars?

The image is (f) $=\left\{f\left(t\right)={t}^{3}+a{t}^{2}+bt+c:tinT\right\}$ ,image is all of $\mathrm{ℝ}$ .

See the step by step solution

## Step 1: Consider the parameters.

The image of a function consists of all the values the function takes in its target space. If f is a function from X to Y, then

$image\left(f\right)=\left\{f\left(x\right):inX\right\}\phantom{\rule{0ex}{0ex}}=\left\{binY:b=f\left(x\right),\mathrm{for}\mathrm{some}x\mathrm{in}X\right\}$

The Kernel of the transformation results in $\left\{f\left(t\right)={t}^{3}+a{t}^{2}+bt+c:tinT\right\}$ , as the reflection is its own inverse.

As the inverse of the reflection exists thus, the image will be $\mathrm{ℝ}$ . The kernel is of dimension $image\left(f\right)=\left\{f\left(t\right)={t}^{3}+a{t}^{2}+bt=c:tinT\right\}$ so, the image will be of $\mathrm{ℝ}$ .

## Step 2: Final answer.

The image is $image\left(f\right)=\left\{f\left(t\right)={t}^{3}+a{t}^{2}+bt+c:tinT\right\}$ , image is all of $\mathrm{ℝ}$ .