Answer: 
Step-by-step explanation:
Add 
on both sides and subtract 
on both sides to leave x's on the left side and independent values on the right.
Solve the fractions.
Convert the mixed fraction 
to an improper fraction. You can do this by multiplying 1 times 35 and adding 2.
Now use the reciprocal (inverse fraction) and multiply on both sides to isolate x.
<span>I note that this problem starts out with "Which is a factor of ... " This implies that you were given several answer choices. If that's the case, it's unfortunate that you haven't shared them.
I thought I'd try finding roots of this function using synthetic division. See below:
f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35
Please use " ^ " to denote exponentiation. Thanks.
Possible zeros of this poly are factors of 35: plus or minus 1, plus or minus 5, plus or minus 7. Use synthetic division; determine whether or not there is a non-zero remainder in each case. If none of these work, form rational divisors from 35 and 6 and try them: 5/6, 7/6, 1/6, etc.
Provided that you have copied down the function
</span>f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35 properly, this approach will eventually turn up 1 or 2 zeros of this poly. Obviously it'd be much easier if you'd check out the possible answers given you with this problem.
By graphing this function, I found that the graph crosses the x-axis at 7/2. There is another root.
Using synth. div. to check whether or not 7/2 is a root:
___________________________
7/2 / 6 -21 -4 24 -35
21 0 -14 35
----------- ------------------------------
6 0 -4 10 0
Because the remainder is zero, 7/2 (or 3.5) is a root of the polynomial. Thus, (x-3.5), or (x-7/2), is a factor.
Step-by-step explanation:
(a) ∫₋ₒₒ°° f(x) dx
We can split this into three integrals:
= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx
Since the function is even (symmetrical about the y-axis), we can further simplify this as:
= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx
The first integral is finite, so it converges.
For the second integral, we can use comparison test.
g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.
We can show that g(x) converges:
∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).
Therefore, the smaller function f(x) also converges.
(b) The width of the intervals is:
Δx = (3 − -3) / 6 = 1
Evaluating the function at the beginning and end of each interval:
f(-3) = e^(-9/2)
f(-2) = e^(-2)
f(-1) = e^(-1/2)
f(0) = 1
f(1) = e^(-1/2)
f(2) = e^(-2)
f(3) = e^(-9/2)
Apply Simpson's rule:
S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]
S ≈ 2.5103
Don't trust me but it's probably the first one. Just using context clues
