PRE CAL HELPPPPPPPPPPPPPPPPPPP?

Question

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Answer 1

1) 1/√(8x-6) We wish to find the values of x for which this function has real values.  

We know that any function of the form 1/g(x) is defined whenever g(x) is not equal to 0.  We also know that the square root of any function is defined when the function is positive. That is √(g(x)) has real values when g(x) ≥ 0.

Using these two facts we wish to find the x for which 8x-6 > 0.

8x - 6 > 0

8x > 6

x > 3/4

2) f(x) = 1/x  and f(x+h) = 1/(x+h)  Plugging into the difference equation we get:

(1/h)[1/(x+h) - 1/x]  We now want to get the two quotients on the right to have the same denominator.  This is done by multiplying 1/(x+h) by x/x and 1/x by (x+h)/(x+h).  This is perfectly legal because it's equivalent to multiplying the entire equation by 1.

(1/h)[x/(x)(x+h) - (x+h)/(x)(x+h)]

(1/h)[-h/(x)(x+h)]

-1/(x)(x+h)

6)  This equation has real values when (x-3)(x+9) ≥ 0

This happens in two cases:

(1) When (x-3) ≥ 0 and (x+9) ≥ 0

or

(2) When (x-3) ≤ 0 and (x+9) ≤ 0  (since two negatives make a positive)

For (1):

x - 3  Ã¢Â‰Â¥0

x  Ã¢Â‰Â¥3

This is equivalent to the half-closed interval [3,∞)

x + 9  Ã¢Â‰Â¥0

x  Ã¢Â‰Â¥ -9

This is equivalent to the half-open interval [-9,∞)

The "and" in (1) indicates an intersection of these two sets.  Thus (1) is equivalent to just [3,∞).  That is when x  Ã¢Â‰Â¥ 3 AND x  Ã¢Â‰Â¥ -9

Now for (2)

x - 3 ≤ 0

x ≤ 3

This is equivalent to (-∞,3]

x + 9 ≤ 0

x  Ã¢Â‰Â¤ -9

This is equivalent to  (-∞,-9]

Once again the "and" in (2) indicates the intersection of  (-∞,3] and  (-∞,-9] which is(-∞,-9]

Looking back at statements (1) and (2) we had (1) OR (2).  This indicates a union so we conclude the domain is any x in (-∞,-9]U[3,∞)

7) f(x) = x^2 + 3x - 6

To find f(x+4) we simply plug in x+4 for x in the original equation.

f(x+4) = (x+4)^2 + 3(x+4) - 6

(x+4)^2 + 12 + 3x - 6

(x+4)^2 + 3x + 6 Expand the first term to get

x^2 + 8x + 16 + 3x + 6

x^2 + 11x + 22