The graph of a function f(x) = ∛x is shown in the picture, and the domain of the function will be all real numbers.
<h3>What is a function?</h3>
It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.
The question is incomplete.
The complete question is in the picture, please refer to the attached picture.
We have a function:
f(x) = ∛x
The domain of the function will be all real numbers.
x ∈(-∞, ∞)
The graph will be curved.
Plug x = 0, y = 0
x = 1, y = 1
The graph is shown in the attached picture.
Thus, the graph of a function f(x) = ∛x is shown in the picture, and the domain of the function will be all real numbers.
Learn more about the function here:
brainly.com/question/5245372
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Answer:
x=
3
/7
or x=
−1
/4
Step-by-step explanation:
Answer:
(-21/24) - (-20/24) = -1/24
<em>Simplest Form of the Equation</em>
(-7/8) - (-5/6) = 0.04166666666
8x6= 48,
-7x6=-42
6x8= 48,
-5x8=40
(-42/48) - (-40/48)
Divide both sides by 2
(-21/24) - (-20/24) = -1/24
A suitable calculator shows the score of
46.0 separates the bottom 26% from the top 74%.
(a) Take the Laplace transform of both sides:
where the transform of 
comes from
![L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)](https://tex.z-dn.net/?f=L%5Bty%27%28t%29%5D%3D-%28L%5By%27%28t%29%5D%29%27%3D-%28sY%28s%29-y%280%29%29%27%3D-Y%28s%29-sY%27%28s%29)
This yields the linear ODE,
Divides both sides by 
:
Find the integrating factor:
Multiply both sides of the ODE by 
:
The left side condenses into the derivative of a product:
Integrate both sides and solve for 
:
(b) Taking the inverse transform of both sides gives
![y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B7t%5E2%7D2%2BC%5C%2CL%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7Be%5E%7Bs%5E2%7D%7D%7Bs%5E3%7D%5Cright%5D)
I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that 
is one solution to the original ODE.
Substitute these into the ODE to see everything checks out:
