1. The range of a function is the set of all values that f can produce for all the x-es in the domain.
2. If we are given the graph, in order to find the range, we project the graph into the y axis. Informally, we draw the "shadow" of the graph into the y axis as in the FIGURE atached.
3. The range is <span>D || {y | −5 ≤ y ≤ −1}</span>
OK. I used my calculator to evaluate sec(85 degrees).
My calculator doesn't have a "sec" button on it.
But I remembered that
sec of an angle = 1 / (cosine of the same angle) .
So I used my calculator to find cos(85), and then I hit the
" 1/x " key, and got 11.474, which I knew to be sec(85).
Answer:
The zeros of the function are;
x = 0 and x = 1
Step-by-step explanation:
The zeroes of the function simply imply that we find the values of x for which the corresponding value of y is 0.
We let y be 0 in the given equation;
y = x^3 - 2x^2 + x
x^3 - 2x^2 + x = 0
We factor out x since x appears in each term on the Left Hand Side;
x ( x^2 - 2x + 1) = 0
This implies that either;
x = 0 or
x^2 - 2x + 1 = 0
We can factorize the equation on the Left Hand Side by determining two numbers whose product is 1 and whose sum is -2. The two numbers by trial and error are found to be -1 and -1. We then replace the middle term by these two numbers;
x^2 -x -x +1 = 0
x(x-1) -1(x-1) = 0
(x-1)(x-1) = 0
x-1 = 0
x = 1
Therefore, the zeros of the function are;
x = 0 and x = 1
The graph of the function is as shown in the attachment below;
