Let f be the function given by f(x)= 2x/(sqrt(x^2 +x +1))
c. Write an equation for each horizontal asymptote of the graph of f.
d. Find the range of f. Use f'(x) to justify your answer.
when x gets really big positive
this is
y = 2 (big) / sqrt(big ^2)
which is y = 2
when x gets really big negative
y = 2 (-big number) / big number
which is y = - 2
well, let's look at the derivative since they say so
[ (x^2+x+1)2 - 2x(.5)(x^2+x+1)^-.5(2x) ]/(x^2+x+1)
yuuk, whose idea was this?
I suggest going to
http://www.mathsisfun.com/data/function-grapher.php
and put in
2 x/(x^2+x+1)^.5
Thank you!
c. Oh, horizontal asymptotes, those sneaky little lines that the graph gets close to but never quite reaches. Now, let's see for this function, as x gets really, really large, what happens? Well, we can simplify the expression a bit by dividing every term by x, but don't worry, I won't leave x out of the equation! So we have (2/x)*(x/(sqrt(x^2 +x +1))). Now, when x goes to infinity, 2/x will become super duper tiny, like a dwarf in Legoland. But the x part in the denominator will grow and dominate everything else. So in the end, the function approaches (1* infinity) / (infinity), which is basically 1/1, or just 1. Therefore, the horizontal asymptote of this function is y = 1.
d. Now, for the range of this jolly function, we can use its derivative, f'(x), to guide us on this journey. First, let's find f'(x), and brace yourself, because it's going to look like a roller coaster ride of calculus. So, f'(x) = (2*(sqrt(x^2 +x +1)) - 2x*(1/2)*(2x + 1))/(x^2 + x + 1). Phew, that was a wild ride, but now we have f'(x) in all its glory. Now, let's try to figure out where f'(x) is positive and where it's negative, just like distinguishing between rainy and sunny days.
If we look closely at the expression, we see that the denominator, x^2 + x + 1, is always positive because it's a quadratic polynomial that never hits 0. So, f'(x) will be positive when the numerator is positive, and negative when the numerator is negative. Now, let's do some detective work with the numerator.
The term (2*(sqrt(x^2 +x +1)) will always be positive. But the term - 2x*(1/2)*(2x + 1) can be both positive and negative. So here's the scoop: when 2x*(1/2)*(2x + 1) is positive, f'(x) will be positive. When it's negative, f'(x) will be negative.
Now, using all this information, we can conclude that the range of the function f will be all real numbers greater than 1, including 1, because as x approaches infinity, f(x) approaches 1. So it's like the range is stretching its arms to include all the bigger numbers. To infinity and beyond, am I right?
Hope that clarifies things and puts a smile on your face.
c. To find the horizontal asymptotes of the function f, we need to examine the behavior of the function as x approaches positive infinity and negative infinity.
As x approaches positive infinity, we can simplify the function f(x) by ignoring the lower order terms in the denominator:
lim(x→∞) (2x/(sqrt(x^2 + x + 1))) = lim(x→∞) (2x/(sqrt(x^2)))
Since the denominator becomes simply x as x approaches infinity, we can rewrite the limit as:
lim(x→∞) (2x/x) = lim(x→∞) 2 = 2
Therefore, the horizontal asymptote of the graph of f is y = 2 as x approaches positive infinity.
Similarly, as x approaches negative infinity:
lim(x→-∞) (2x/(sqrt(x^2 + x + 1))) = lim(x→-∞) (2x/(sqrt(x^2)))
Again, the denominator becomes simply x as x approaches negative infinity:
lim(x→-∞) (2x/x) = lim(x→-∞) 2 = 2
Therefore, the horizontal asymptote of the graph of f is y = 2 as x approaches negative infinity.
d. To find the range of the function f, we can use the derivative f'(x) and analyze its behavior.
First, let's find the derivative of f(x):
f'(x) = d/dx (2x/(sqrt(x^2 + x + 1)))
= (2(sqrt(x^2 + x + 1)) - 2x(1/2)(x^2 + x + 1)^(-1/2))/(x^2 + x + 1)
Simplifying the above expression, we get:
f'(x) = (2(x^2 + x + 1) - x(sqrt(x^2 + x + 1)))/(x^2 + x + 1)
= (3x + 2)/(x^2 + x + 1)
The denominator x^2 + x + 1 is always positive, so we only need to consider the behavior of the numerator 3x + 2.
By setting 3x + 2 equal to 0 and solving for x, we find:
3x + 2 = 0
3x = -2
x = -2/3
The numerator 3x + 2 is always positive for x > -2/3, and always negative for x < -2/3.
Therefore, the range of the function f is all real numbers except for the values below -2/3.
In interval notation, the range of f is (-∞, -2/3) ∪ (-2/3, ∞).
To find the horizontal asymptotes of the function f(x), we need to analyze the behavior of the function as x approaches positive or negative infinity.
a. Horizontal Asymptote: Let's start by finding the horizontal asymptotes of the function.
The general formula for determining horizontal asymptotes is:
1. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.
2. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.
3. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
For function f(x) = 2x/√(x^2 + x + 1):
1. The degree of the numerator is 1, and the degree of the denominator is 2. Therefore, the degree of the numerator is less than the degree of the denominator.
So, according to the first case, the horizontal asymptote is y = 0.
b. Range of f(x): The range of a function represents the set of all possible y-values that the function can attain.
To find the range, we can analyze the behavior of the derivative of the function f'(x), as x approaches positive or negative infinity.
Let's find the derivative f'(x) first:
f(x) = 2x/√(x^2 + x + 1)
Using the quotient rule, we can find the derivative as follows:
f'(x) = [2(√(x^2 + x + 1)) - 2x(1/2)(x^2 + x + 1)^(-1/2)] / (x^2 + x + 1)
Simplifying further, f'(x) is equal to:
f'(x) = [2√(x^2 + x + 1) - x/√(x^2 + x + 1)] / (x^2 + x + 1)
Next, we will analyze the behavior of f'(x) as x approaches positive or negative infinity.
1. As x approaches positive infinity:
- The numerator 2√(x^2 + x + 1) grows without bound.
- The denominator (x^2 + x + 1) also grows without bound.
- Therefore, both the numerator and denominator approach infinity, but the numerator grows faster.
- As a result, the limit of f'(x) as x approaches positive infinity is positive infinity.
2. As x approaches negative infinity:
- Similarly, both the numerator and denominator grow without bound.
- However, as x approaches negative infinity, the term x/√(x^2 + x + 1) approaches -1.
- Therefore, the limit of f'(x) as x approaches negative infinity is -1.
To summarize:
- As x approaches positive infinity, f'(x) approaches positive infinity.
- As x approaches negative infinity, f'(x) approaches -1.
By analyzing the behavior of f'(x), we can conclude the range of f(x):
- The function is continuously increasing without bound from negative infinity to positive infinity.
- Therefore, the range of f(x) is (-∞, +∞) or all real numbers.
c. Equation for each horizontal asymptote: The equation for the horizontal asymptote was found in part a as y = 0.
In conclusion:
a. The equation for the horizontal asymptote of the graph of f is y = 0.
b. The range of f is (-∞, +∞) or all real numbers.