Problem 18

Question

The graph of f is given. (a) Why is f one-to-one? (b) What are the domain and range of f^{-1} ? (c) What is the value of f^{-1} (2) ? (d) Estimate the value of f^{-1} (0) .

Accepted Answer

all right. So here we have a rough sketch of the graph of F, and it&#x27;s 1 to 1 because every X value has only one y value and every y value has only one X value. And graphically, we can say it passes the horizontal line test as well as a vertical line test. Okay, now we went to find the domain and range of F inverse. So first of all, let&#x27;s find the domain and range of F. So the domain of F the X values that goes through our negative 3 to 3 on the range of death, the Y values that goes through our negative 123 So the domain of F inverse will be the range of F, and the range of FM burst will be the domain of F. So we&#x27;re going to switch these and the domain of F inverse will be from negative 123 and the range will be from negative 3 to 3. When you have an inverse outputs become inputs, inputs become outputs. All right. Now let&#x27;s find the value of f inverse of to so effin versus to let&#x27;s say it equals X, then that means that f of X equals two, the output of X inverse will be the inverse will be the input of F inputs and outputs are switched. So what? X value appears to have a Y value of two. Looking at the graph, it looks like X equals zero has a Y value of two. So f inverse of two is zero. And finally, let&#x27;s estimate the value of F inverse of zero. So similarly, let&#x27;s just say that it&#x27;s X, so that means that f of X equals zero. So we&#x27;re estimating the value of the point where there&#x27;s a wide coordinate of zero on the graph. Now you really should look at the graph in the book, not my graph, because mine is just a rough sketch. And if you look at the graph in the book, you see that it passes through a height of zero at about negative 1.5 negative 1.6 somewhere in there, so f inverse of zero is approximately negative. 1.6

Problem

The formula $ C = \frac{5}{9} (F - 32) $ , where …

Problem 18 Medium Difficulty

The graph of $ f $ is given.
(a) Why is $ f $ one-to-one?
(b) What are the domain and range of $ f^{-1} $ ?
(c) What is the value of $ f^{-1} (2) $?
(d) Estimate the value of $ f^{-1} (0) $.

Answer

a) $f$ is $1-$ to-l because there is no horizontal line that intersects the graph more
than once.
b) The domain of $f^{-1}$ is $[-1,3]$ and the range is $[-3,3]$
c) $f^{-1}(2)=0$
d) $f^{-1}(0) \approx-1.7$

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a) $f$ is $1-$ to-l because there is no horizontal line that intersects the graph morethan once.b) The domain of $f^{-1}$ is $[-1,3]$ and the range is $[-3,3]$c) $f^{-1}(2)=0$d) $f^{-1}(0) \approx-1.7$

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Catherine R.

Heather Z.

Caleb E.

Joseph L.

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Vector Basics - Example 1

Kayleah T.

Caleb E.

Samuel H.

Joseph L.

a) $f$ is $1-$ to-l because there is no horizontal line that intersects the graph more
than once.
b) The domain of $f^{-1}$ is $[-1,3]$ and the range is $[-3,3]$
c) $f^{-1}(2)=0$
d) $f^{-1}(0) \approx-1.7$