The answer is 78% that was rounded from 77.78%
Answer:
For a function y = f(x), the range is the set of all the possible values of y.
In the question you wrote:
y = secx - 2
This can be interpreted as:
y = sec(x - 2)
or
y = sec(x) - 2
So let's see each case (these are kinda the same)
If the function is:
y = sec(x - 2)
Firs remember that:
sec(x) = 1/cos(x)
then we can rewrite:
y = 1/cos(x - 2)
notice that the function cos(x) has the range -1 ≤ y ≤ 1
Then for the two extremes we have:
y = 1/1 = 1
y = 1/-1 = -1
Notice that for:
y = 1/cos(x - 2)
y can never be in the range -1 < x < 1
As the denominator cant be larger, in absolute value, than 1.
Then we can conclude that the range is all reals except the interval:
-1 < y < 1
If instead the function was:
y = sec(x) - 2
y = 1/cos(x) - 2
Then with the same reasoning, the range will be the set of all real values except:
-1 - 2 < y < 1 - 2
-3 < y < -1
This is how you would graph y=5x+1
Answer: √(f^2 - 2f + 1) = a
Step-by-step explanation:
We have that:
f^2+g^4-2f+1 = a^2+(g^2)^2
And we want to find the value of a, so we should isolate it.
f^2 + g^4 - 2f + 1 = a^2 + g^(2*2) = a^2 + g^4
where i used that (x^y)^z = x^(y*z)
We can remove the term g^4 in both sides of the equation and get:
f^2 - 2f + 1 = a^2
now we can apply the square rooth to both sides and get
√(f^2 - 2f + 1) = a
So we just found the value of a.
