# Assume the function has an inverse. without solving for the inverse find the indicated function values:

###### Question:

## Answers

F(x) = x³ + 4x - 1

The domain of a function is the range of its inverse and the range of a function is the domain of its inverse. Thus, to find f⁻¹(x), we put the given value of x into f(x) of the original equation.

a) -1 = x³ + 4x - 1

x = 0

f⁻¹(-1) = 0

b) 4 = x³ + 4x - 1

x = 0.25

f⁻¹(4) = 0.25

486

Step-by-step explanation:

I just did 6/2 and got three and that was the rate of change so I multiplied 3 to 2 to get 6 and then multiplied 3 again to get 18 and kept going until I got to 162 and multiplied that to 3 to get 486.

486

Step-by-step explanation:

it multiplies by 3 each time

answer:

your answer is:

step-by-step explanation:

actually, i can't answer because you have not included the function! include the function, and then i can answer. you!

Explanation:

A function f has an absolute minimum at x = c if f(c) is the smallest function value on the entire domain of f, whereas f has a local minimum at c if f(c) is the smallest function value when x is near c.

___

That pretty much covers it.

__

The attached graph shows a function with a local minimum near x = -0.432 and an absolute minimum (which is also a local minimum) near x = 2.141.

[tex]Explain the difference between an absolute minimum and a local minimum. a function f has an absolute[/tex]

f(0) = 2

f(x) = 0 at x=-1

Step-by-step explanation:

f(0) is the value of the graph when x=0

f(0) = 2

f(x) = 0 is when the graph is 0 (the y value is zero)

f(x) is zero when x=-1

If x = 0, then f(0) = 2.25

If f(x) = 0, then x = -1

if x = 0, then f(0) = 2.25

if f(x) = 0, then x = -1

Step-by-step explanation:

I got it correct on edge

roots? I'm pretty sure

Step-by-step explanation:

Absolute minimum is where, in the infinite domain of f(x), the least value the function is the absolute minimum.

However, the local minimum is only a specific interval on the domain of f(x) where the lowest value is, but only in that interval.

So, usually, this comes up when you're dealing with 1st and 2nd derivatives. The first derivative tells you where local max and mins are, where the 2nd derivative tells you if the concavity is up or down.