For question (1), you haven't used any parentheses, but based on some of your later problems, I'm going to assume that you indeed meant the function, y(x) = (x - 1) / (x - 4). To figure out where this function is greater than zero, we certainly need to find out where it is equal to zero. Accordingly, we set y(x) equal to zero:
y(x) = (x - 1) / (x - 4) = 0
Right away, we recognize that x cannot equal 4, or we'll be dividing by zero. We should also recognize that the only zero of the function is x = 1. These two points, x = 1 and x = 4, are the ones we're interested in. If a function changes sign from negative to positive or vice versa, it must do so at either a zero or a point where the value of the function is undefined. Hence, to the left of x = 1, the function will have some sign, between x = 1 and x = 4, it will have some sign, and to the right of x = 4 it will have some sign. We therefore need only to pick a single point in each region to determine the sign of each region as a whole. Let's do ourselves a favor and pick easy values, like x = 0, x = 2, and x = 5. We plug them in and check the sign of each:
y(0) = (0 - 1) / (0 - 4)
y(0) = -1 / -4
y(0) = 1 / 4
This is positive, so for x ≤ 1, y(x) ≥ 0.
y(2) = (2 - 1) / (2 - 4)
y(2) = 1 / -2
y(2) = -1 / 2
This is negative, so for 1 < x < 4, y(x) < 0.
y(5) = (5 - 1) / (5 - 4)
y(5) = 4 / 1
y(5) = 4
This is positive, so for x ≥ 4, y(x) ≥ 0.
Combining the two positive regions we found, we can say that for x ≤ 1 or x > 4, y(x) ≥ 0. We use x > 4 rather than x ≥ 4, since at x = 4, y(x) is undefined.
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On to problem (2). If we want the graph of y = √x to shift to the right, we want the y-values to be smaller for a given x-value. Accordingly, the delayed function will be of the form:
y(x) = √(x - a)
If the shift is to be 4 units in x, then a = 4, and we write:
y(x) = √(x - 4)
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Problem (3) is asking us about the range of a function. Recall that the domain of a function is the set of x-values it can possibly take, while the range of a function is the set of y-values it can possibly take. For the function:
y(x) = 6 cos^2(x)
...let's step back and consider the simpler case of:
y(x) = cos(x)
You know that the sine and cosine functions have a domain of all real numbers (so any value of x), and that they have a range of -1 ≤ x ≤ 1. This is just how they are defined.
What happens if we square the cosine term? Pick any number on the interval -1 ≤ x ≤ 1 (the original range), and square it. You get a number between 0 and 1. Hence, the range of the squared function is 0 ≤ x ≤ 1.
What happens if we also multiply by 6? This is an easy one. The range is multiplied by 6, since for any y-value, we're simply multiplying it by 6. Hence, the range of the function in the question is 0 ≤ x ≤ 6.
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Problem (4) is a very simple one. It's just asking, what do we have to do to:
y = 5x
...to get it to be:
y = 65x
The only difference is the value of the leading coefficient, so the "transformation" between the two equations must just be to multiply by some number. Since 65 / 5 = 13, the answer is to multiply by 13.
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For problem (5), I'm going to assume that you meant to use parentheses:
f(x) = (5x^2 + 2x - 3) / (2x^2 - 7x + 4)
The asymptotic behavior of a function like this just means the behavior of the function as x takes extremely large values (negative infinity or positive infinity). If you compare the graphs of f(x) = x and g(x) = x^2, you see that g(x) increases much faster than f(x).
Let's consider two functions:
g(x) = x^2
h(x) = x^2 + x
Input these two functions to your graphing calculator, and plot them simultaneously. You see that for small x-values, the two functions take rather different values. Now choose a bigger window. You should see that for very large x-values, the two functions converge to closer and closer values. Hence, the x term becomes less and less significant for higher and higher x-values.
We can generalize this by saying that, for very large values of x, in considering polynomial expressions of the form:
f(x) = ax^n + bx^(n - 1) + cx^(n - 2) + ... + z
...only the highest-order term (ax^n) is significant.
So to return to the original problem, let's consider the numerator as a function and the denominator as a function. Both are polynomials, and most importantly, both have the same order (they're both quadratic). This means that, for very large values of x, they tend to converge on one another. Thus, dividing one by the other will give us values of f(x) that get asymptotically closer and closer to 1.
...or at least, that would be the case if not for those leading coefficients. If we ignore all the terms except for the x^2 terms, we have:
f(x) = 5x^2 / 2x^2
We can simplify this:
f(x) = 5 / 2
This is the value of the horizontal asymptote for very large x-values (as x approaches positive or negative infinity). This method of ignoring all but the highest-order x terms, and then simplifying, is the method you should use, and hopefully now you can see why. Note that if the order of the numerator and denominator are not equal, you'll get different asymptotic behavior. For example, if the numerator had been 5x^3 instead of 5x^2, it would have simplified to:
f(x) = 5x / 2
And you would have had a so-called "slant asymptote," not a horizontal one. If the numerator had been 5x^4 or higher, you wouldn't have an asymptote in the usual sense, since the function would simplify to a quadratic for large values of x.
If, on the other hand, the numerator had been of any order lower than the denominator, you'd arrive at something like:
f(x) = 5 / 2x
or
f(x) = 5 / 2x^2
You know offhand that expressions like these tend to zero as the value of x gets bigger and bigger, so these, too have horizontal asymptotes (at f(x) = 0).
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For the next problem, you have:
e^(2x) = 5.8
You may recall the following identities:
ln(e^(f(x)) = f(x)
e^ln(f(x)) = f(x)
This allows us to take the natural logarithm of both sides of the equation, obtaining:
ln(e^(2x)) = ln 5.8
2x = ln 5.8
x = (ln 5.8) / 2
Next, recall the property of following property of logarithms:
k ln f(x) = ln ((f(x))^k)
In our case, k = 1 / 2, so we can raise 5.8 to the one-half power (take the square root):
x = ln √5.8
How you solve this problem is up to you. I'd use a calculator, personally.
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The equation:
2 - sin(x) = 6
...can be solved simply:
2 = 6 + sin(x)
sin(x) = -4
x = sin^-1(-4)
Remember that sin^-1(x) is arcsine, not one divided by sine. In our case, the expression means the angle x for which sin(x) = -4. But right there you should notice that there's a problem: recall from one of the previous problems that the range of the sine and cosine functions is -1 ≤ x ≤ 1. There is no value of x for which sin(x) = -4; hence, there is no solution to this equation.
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The next problem is very similar to the first problem. If you apply the same methods, you'll find that y(x) ≥ 0 when x ≤ -8 or x > 3. As in the first problem, we say x > 3 instead of x ≥ 3, since for x = 3, y(x) is undefined.
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The last problem is simply asking you to compose two functions. Recall:
f(x) ○ g(x) = f(g(x))
g(x) ○ f(x) = g(f(x))
The two are not generally the same, so pay attention to the order. In our case:
f(x) = 3x^2 + 4
g(x) = x - 7
So to compose f with g, we simply plug g(x) in for x in f(x):
f ○ g = f(g(x))
f ○ g = 3(x - 7)^2 + 4
f ○ g = 3(x^2 - 14x + 49) + 4
f ○ g = 3x^2 - 42x + 151
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If you're not clear on any of these solutions, or have an additional question, just update your question and let me know.