it goes up 0.8 every time
Answer:
x > - 3
Step-by-step explanation:
x + 2
> -
First turn 2
into an fraction since its easier. So it becomes 
x +
> - 
<u> - </u>
<u> - </u>
<u> </u> Do inverse operations, so subtract
from both sides.
x > -
Combine -
with -
, which gives us - 
x > - 3 Since we know that 6 divided by 2 is 3.
This would not allow me to attach an image so I am going to type a visual representation f what the graph would look like below)
<-------------O--------------------------------------------------->
-5 -4 -3 -2 -1 0 1 2 3 4 5
So, the number line would be a solid line rather than dashed (but I was not able to attach an image so I kind of had to improvise). When using ≥ ≤ sings the circle of the point is closed. When using < and > signs, the circle of the point is open. Thus, in this scenario since he sign is >, then the circle of the point is open. The direction of the arrow and ray for this equality is going to the right because the answer is that x is greater than - 3, Therefore any values greater than - 3 are a possibility which is why the arrow goes on indefinitely.
![\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x} \\\\\\ (-2)^3=x\implies \boxed{-8=x}\\\\ -------------------------------\\\\ 0=\sqrt[3]{x}\implies \boxed{0=x}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dx%2B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B3%5Cleft%28%5Cfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%20%20%5Cright%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B%5Ccfrac%7B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Csqrt%5B3%5D%7Bx%7D%2B2%5Cimplies%20-2%3D%5Csqrt%5B3%5D%7Bx%7D%0A%5C%5C%5C%5C%5C%5C%0A%28-2%29%5E3%3Dx%5Cimplies%20%5Cboxed%7B-8%3Dx%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A0%3D%5Csqrt%5B3%5D%7Bx%7D%5Cimplies%20%5Cboxed%7B0%3Dx%7D)
now, f(0) = 0, and f(-8) is an imaginary value or no real value.
now, f(-10) will also give us an imaginary value
and f(1) = 4
so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.
however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.
the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).