Solve log9^5 = 7 - 3x by graphing
2 answers:
You might be interested in
9514 1404 393
Explanation:
1. End behavior is the behavior of the function when the value of the independent variable gets large (or otherwise approaches the end of the domain). There are generally four kinds of end behavior:
- the function approaches a constant (horizontal asymptote)
- the function approaches a function (slant asymptote, for example)
- the function oscillates between two of the above end behaviors
- the function tends toward +∞ or -∞
Of these, behavior 2 will ultimately look like one of the others.
For polynomials, the function will always approach ±∞ as the independent variable approaches ±∞. Whether the signs of the infinities agree or not depends on the even/odd degree of the polynomial, and the sign of its leading coefficient.
For exponential functions, the end behavior is a horizontal asymptote in one direction and a tending toward ±∞ in the other direction.
For trig functions sine and cosine, the end behavior is the same as the "middle" behavior: the function oscillates between two extreme values.
For rational functions (ratios of polynomials), the end behavior will depend on the difference in degree between numerator and denominator. If the degree of the denominator is greater than or equal to that of the numerator, the function will have a horizontal asymptote. If the degree of the numerator is greater, then the end behavior will asymptotically approach the quotient of the two functions—often a "slant asymptote".
__
2. A polynomial inequality written in the form f(x) ≥ 0, or f(x) > 0, will be solved by first identifying the real zeros of the function f(x), including the multiplicity of each. For positive values of x greater than the largest zero, the sign of the function will match the sign of the leading coefficient. The sign will change at each zero that has odd multiplicity, so one can work right to left to identify the sign of the function in each interval between odd-multiplicity zeros.
The value of the function will be zero at each even-multiplicity zero, but will not change sign there. Obviously, the zero at that point will not be included in the solution interval if the inequality is f(x) > 0, but will be if it is f(x) ≥ 0. Once the sign of the function is identified in each interval, the solution to the inequality becomes evident.
As a check on your work, you will notice that the sign of the function for x > max(zeros) will be the same as the sign of the function for x < min(zeros) if the function is of even degree; otherwise, the signs will be different.
The solution to a polynomial inequality is a set of intervals on the real number line. The solution to a polynomial equation is a set of points, which may be in the complex plane.
__
3. A composite function is a function of a function, or a function of a composite function. For example f(g(x)) is a composite function. The composition can be written using either of the equivalent forms ...
It can be easy to confuse an improperly written composition operator with a multiplication symbol, so the form f(g(x)) is preferred when the appropriate typography is not available.
When simplifying the form of a composition, the Order of Operations applies. That is, inner parenthetical expressions are evaluated (or simplified) first. As with any function, the argument of the function is substituted wherever the independent variable appears.
For example, in computing the value f(g(2)), first the value of g(2) is determined, then that value is used as the argument of the function f. The same is true of other arguments, whether a single variable, or some complicated expression, or even another composition.
Note that the expression f(g(x)) is written as the composition shown above. The expression g(f(x)) would be written using the composition operator with g on the left of it, and f on the right of it:
That is, with respect to the argument of the composition, the functions in a composition expression are right-associative. For example, ...
for h(x)=2x+3, g(x)=x^2, f(x)=x-2 we can evaluate f(g(h(x)) as follows:
f(g(h(x)) = f(g(2x+3) = f((2x+3)^2) = (2x+3)^2 -2
It should be obvious that g(h(f(x)) will have a different result.
g(h(f(x)) = g(h(x-2)) = g(2(x-2)+3) = (2(x-2)+3)^2