Gregory Jones

2022-01-12

Does the function $f\left(x\right)=\frac{1}{x}$ have an inverse function?

otoplilp1

Beginner2022-01-13Added 41 answers

Step 1

The horizontal line test only tells us the function is injective. Anyway, before we start talking about injectivity/surjectivity/biijectivity, we should always specify domains and target spaces.

The function

is injective, but not surjective.

The function

is bijective, and

The function

is injective but not surjective.

The function

is bijective, and

And so on. But note that typically, if we have a function

Andrew Reyes

Beginner2022-01-14Added 24 answers

Step 1

The horizontal line test you mention only checks if f is injective. The correct test would be: every horizontal line crosses the graph exactly once. (This is assuming the codomain is$\mathbb{R}$ . Otherwise, the horizontal lines should be restricted to the appropriate codomain.)

Indeed, f has an inverse when considered as a function from$\frac{\mathbb{R}}{\left\{0\right\}}$ to $\frac{\mathbb{R}}{\left\{0\right\}}$ (And not when the codomain is considered to be $\mathbb{R}$ )

But there's a more general phenomenon to be considered here: Suppose$f:A\to B$ is a function. The codomain is not so ''inherent'' to the function as much as its image (range). You could always just take a superset of the codomain and still have the ''same function''.In particular, if $f:A\to B$ is an injective function, then the ''restriction'' $f:A\to f\left(A\right)$ is a bijection and it has an inverse. (f(A) denotes the image of f.)

In this sense, you only need an injection to have an inverse, which is what the original horizontal line test says.

The horizontal line test you mention only checks if f is injective. The correct test would be: every horizontal line crosses the graph exactly once. (This is assuming the codomain is

Indeed, f has an inverse when considered as a function from

But there's a more general phenomenon to be considered here: Suppose

In this sense, you only need an injection to have an inverse, which is what the original horizontal line test says.

alenahelenash

Expert2022-01-24Added 556 answers

Step 1
First of all, the domain of the function f described by
$f(x)=\frac{1}{x}$ is $\frac{\mathbb{R}}{\{0\}}$
(you can plug anything but 0 into this function). The range of the function is also $\frac{\mathbb{R}}{\{0\}}$ because the equation
$y=\frac{1}{x}$
is solvable iff
$y\ne 0$
Because the solution of $y=\frac{1}{x}$ is unique for any $y\ne 0$ , the function is injective (which corresponds to the horizontal line check). You could also see this by
$\frac{1}{x}=\frac{1}{y}\iff 1=\frac{x}{y}\iff x=y$
because $y\ne 0$ and $x\ne 0$
Any injective function has an inverse if you reduce the codomain (which becomes the domain of the inverse) so that the function is also surjective (just make the range of the function the codomain). Therefore
$f:\frac{\mathbb{R}}{\{0\}}\to \mathbb{R}$
$f(x)=\frac{1}{x}$
would not be invertible - but if we manipulate the codomain and change it to
$f:\frac{\mathbb{R}}{\{0\}}\to \frac{\mathbb{R}}{\{0\}}$
$f(x)=\frac{1}{x}$
is invertible and
$y=\frac{1}{x}\iff x=\frac{1}{y}$
tells us that the inverse of f is f itself.

What is 4 over 16 simplified to?

Using impulse formula, derive p_1 i+p_2 i=p_1 f+p_2 f formula where i-is initial state, f-is the final state after collision. Thanks

Averaging Newton's Method and Halley's Method.

Are these two methods identical? / Do they return the same result per iteration?Difference between Newton's method and Gauss-Newton method

Prove that Newton's Method applied to $f(x)=ax+b$ converges in one step? Would it be because the derivative of $f(x)$ is simply $a$?

simlifier tan(arcsin(x))

$dy/dx=\mathrm{sinh}(x)$ A tangent line through the origin has equation $y=mx$. If it meets the graph at $x=a$, then $ma=\mathrm{cosh}(a)$ and $m=\mathrm{sinh}(a)$. Therefore, $a\mathrm{sinh}(a)=\mathrm{cosh}(a)$.

Use Newton's Method to solve for $a$Using Newton's method below:

${x}_{n+1}={x}_{n}-\frac{f({x}_{n})}{{f}^{\prime}({x}_{0})}$

using this chord formula where the chord length $c$ is $1$ cm:

$c=2r\mathrm{sin}\frac{\theta}{2}$

supposing the radius is $1.1$ cm and the angle $\theta $ is unknown, show the iterative Newton's Method equation you would use to find an approximate value for $\theta $ in the context of this problem (using the appropriate function and derivative).Estimate the number of iterations of Newton's method needed to find a root of $f(x)=\mathrm{cos}(x)-x$ to within ${10}^{-100}$.

Why the bisection method is slower than Newton's method from a complexity point of view?

To find approximate $\sqrt{a}$ we can use Newton's method to approximately solve the equation ${x}^{2}-a=0$ for $x$, starting from some rational ${x}_{0}$.

Newton's method in general is only locally convergent, so we have to be careful with initialization.

Show that in this case, the method always converges to something if ${x}_{0}\ne 0$.Consider the function $f(x,y)=5{x}^{2}+5{y}^{2}-xy-11x+11y+11$. Consider applying Newton's Method for minimizing f. How many iterations are needed to reach the global minimum point?

1. If you weigh 140 lbs on Earth, what is your mass in kilograms

2.Using the answer from problem I (hopefully it's correct), determine your weight in Newton's if you were on the moon. Yes, you have to look up something to complete this problem.

3. You measured the mass of a rock to be 355g. What is its weight?

4.You are now holding the rock in your hand from problem 3. What force is the rock on your hand? How much force do you need to exert on the rock to hold in stationary.A solution containing 5.24 mg/100 mL of A (335 g/mol) has

**a transmittance of 55.2% in a 1.50-cm cell at 425 nm**.find molar absorbtivityFind the cube root of 9, using the Newton's method.