The domain is going to be all ur x values...so the domain is { -4,-1,1,4 }....the range will be all ur y values...so the range is { -1,1,4} ...the reason there is only 3 numbers is because if the numbers repeat, u only have to write them once.
A function will not have any repeating x values....they can have repeating y values, just not the x ones. So if all ur x values are different, then it is a function. Therefore, this is a function.
<span>sqrt(3x+7)=x-1 </span>One solution was found : <span> x = 6
</span>Radical Equation entered :
<span> √3x+7 = x-1
</span>
Step by step solution :<span>Step 1 :</span>Isolate the square root on the left hand side :
Radical already isolated
<span> √3x+7 = x-1
</span>
<span>Step 2 :</span>Eliminate the radical on the left hand side :
Raise both sides to the second power
<span> (√3x+7)2 = (x-1)2
</span> After squaring
<span> 3x+7 = x2-2x+1
</span>
<span>Step 3 :</span>Solve the quadratic equation :
Rearranged equation
<span> x2 - 5x -6 = 0
</span>
This equation has two rational roots:
<span> {x1, x2}={6, -1}
</span>
<span>Step 4 :</span>Check that the first solution is correct :
Original equation
<span> √3x+7 = x-1
</span> Plug in 6 for x
<span> √3•(6)+7 = (6)-1
</span> Simplify
<span> √25 = 5
</span> Solution checks !!
Solution is:
<span> x = 6
</span>
<span>Step 5 :</span>Check that the second solution is correct :
Original equation
<span> √3x+7 = x-1
</span> Plug in -1 for x
<span> √3•(-1)+7 = (-1)-1
</span> Simplify
<span> √4 = -2
</span> Solution does not check
2 ≠ -2
One solution was found : <span> x = 6</span>
Answer:
23dddsertyh HD ddderfejqjqi2828384 urn ndndnnd
Step-by-step explanation:
jdjejjwkwiwjebdbj djdjrbrnrr
Answer:
A: 3√x^4yx^2
Step-by-step explanation:
Hopefully this helps.
Answer: 2x + h - 2
===============================================
Work Shown:
Part 1
![f(x) = x^2 - 2x + 8\\\\f(x+h) = (x+h)^2 - 2(x+h) + 8\\\\f(x+h) = (x+h)(x+h) - 2(x+h) + 8\\\\f(x+h) = x(x+h)+h(x+h) - 2(x+h) + 8\\\\f(x+h) = x^2+xh+xh+h^2 - 2x-2h + 8\\\\f(x+h) = x^2+2xh+h^2 - 2x-2h + 8\\\\](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E2%20-%202x%20%2B%208%5C%5C%5C%5Cf%28x%2Bh%29%20%3D%20%28x%2Bh%29%5E2%20-%202%28x%2Bh%29%20%2B%208%5C%5C%5C%5Cf%28x%2Bh%29%20%3D%20%28x%2Bh%29%28x%2Bh%29%20-%202%28x%2Bh%29%20%2B%208%5C%5C%5C%5Cf%28x%2Bh%29%20%3D%20x%28x%2Bh%29%2Bh%28x%2Bh%29%20-%202%28x%2Bh%29%20%2B%208%5C%5C%5C%5Cf%28x%2Bh%29%20%3D%20x%5E2%2Bxh%2Bxh%2Bh%5E2%20-%202x-2h%20%2B%208%5C%5C%5C%5Cf%28x%2Bh%29%20%3D%20x%5E2%2B2xh%2Bh%5E2%20-%202x-2h%20%2B%208%5C%5C%5C%5C)
Part 2
![\frac{f(x+h)-f(x)}{h} = \frac{(x^2+2xh+h^2 - 2x-2h + 8) - (x^2-2x+8)}{h}\\\\\frac{f(x+h)-f(x)}{h} = \frac{x^2+2xh+h^2 - 2x-2h + 8 - x^2+2x-8}{h}\\\\\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2-2h}{h}\\\\\frac{f(x+h)-f(x)}{h} = \frac{h(2x+h-2)}{h}\\\\\frac{f(x+h)-f(x)}{h} =2x+h-2\\\\](https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20%3D%20%5Cfrac%7B%28x%5E2%2B2xh%2Bh%5E2%20-%202x-2h%20%2B%208%29%20-%20%28x%5E2-2x%2B8%29%7D%7Bh%7D%5C%5C%5C%5C%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20%3D%20%5Cfrac%7Bx%5E2%2B2xh%2Bh%5E2%20-%202x-2h%20%2B%208%20-%20x%5E2%2B2x-8%7D%7Bh%7D%5C%5C%5C%5C%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20%3D%20%5Cfrac%7B2xh%2Bh%5E2-2h%7D%7Bh%7D%5C%5C%5C%5C%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20%3D%20%5Cfrac%7Bh%282x%2Bh-2%29%7D%7Bh%7D%5C%5C%5C%5C%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20%3D2x%2Bh-2%5C%5C%5C%5C)
-------------
In part 1, I replaced every x with x+h. Then I expanded things out using the distributive property. This is to figure out what f(x+h) is equal to.
In part 2, I then computed the difference quotient. Notice how basically all of the terms in the original f(x) cancel out. Leaving nothing but h terms behind. We factor out h to cancel it with the denominator.