Horizontal asymptote for f(x)=10x/5x^2+1
Using calculus, df/dx=10/(5x^2+1)-20x/(5x^2+1)^3(10x)=0
10=2/(5x^2+1)
25x^2+5=1
x^2=-4/25
x has no real solutions
Thank you for the help.
To find the horizontal asymptote of the function f(x) = (10x) / (5x^2 + 1), you need to check what happens to the function as x approaches positive or negative infinity.
1. As x approaches positive infinity, the highest order term in the denominator (5x^2) becomes the dominant term. Therefore, the function can be approximated as f(x) ≈ (10x) / (5x^2) = 2/x. As x gets larger and larger, the value of f(x) gets closer and closer to zero.
2. Similarly, as x approaches negative infinity, the highest order term in the denominator (-5x^2) becomes the dominant term. The function can be approximated as f(x) ≈ (10x) / (-5x^2) = -2/x. As x becomes more and more negative, the value of f(x) approaches zero again.
Based on these observations, we can conclude that the horizontal asymptote of the function f(x) = (10x) / (5x^2 + 1) is y = 0, also known as the x-axis.
To find the horizontal asymptote of a function, we need to analyze what happens to the function as x approaches positive infinity or negative infinity.
In this case, we have the function f(x) = (10x)/(5x^2 + 1).
As x approaches positive infinity, the x^2 term in the denominator becomes dominant relative to the x term. This is because as x gets larger and larger, x^2 increases much faster than x. So, we can ignore the x term in the denominator and focus on the x^2 term.
Therefore, as x approaches positive infinity, the function can be simplified as f(x) ≈ (10x)/(5x^2) = 2/x.
Now, as x gets larger and larger, the value of 2/x approaches 0. So, the horizontal asymptote is y = 0 as x approaches positive infinity.
Similarly, as x approaches negative infinity, the x^2 term in the denominator becomes dominant. So, we can ignore the x term and simplify the function as f(x) ≈ (10x)/(-5x^2) = -2/x.
Again, as x gets smaller and smaller (approaches negative infinity), the value of -2/x approaches 0. So, the horizontal asymptote is y = 0 as x approaches negative infinity.
Therefore, the horizontal asymptote for the function f(x) = (10x)/(5x^2 + 1) is y = 0.