Can someone help me with these.
1.Name any intervals on which f(x)=2x-3/x-2 is increasing or decreasing.
2. Find any horizontal asymptotes of f(x)=10/x+3.
Are you in Calculus?
Did you mean ...
f(x) = (2x-3)/(x-2) ?
then f'(x) = ((x-2)(2) - (2x-3)(1) )/(x-2)^2
= (2x-4 - 2x+3)/x-2)^2
= -1/(x-2)^2
It is easy to see that this expression will be negative for all values of x, x ≠ 2
What have you learned about a function when its derivative is negative?
2. Again, I am sure you meant
f(x) = 10/(x+3)
Get a "feel" for numbers, as x --> +∞ , the denominator becomes +∞,
so 10/huge ---> 0 but would be slightly above the x-axis
as x --> -∞ , the denominator becomes
10/(-huge) --->0 but below the x-axis, (negative)
Can you draw your conclusion from that?
I am in pre-calculus, and my teacher barely explains the content...
As for question 1, there is no brackets in 2x-3 or x-2, I am not sure if it will make a difference to it
I have to rely on youtube and jiksha >.<
I am sure that in your book the 2x-3 is the numerator and the x-2 is the denominator.
In that case you HAVE to use brackets to force that order of operation.
e.g. if x = 6, the way I read it, the value would be
(2(6) -3)/(6-2) = 9/4
subbing x=3 into your typed expression gives me
2(6) - 3/6 - 2
= 12 - 1/2 - 2
= 9 1/2
so, of course it matters.
1. To determine the intervals on which the function f(x) = 2x - 3/(x - 2) is increasing or decreasing, you can follow these steps:
Step 1: Identify the critical points by finding the values of x where the derivative of the function is equal to zero or undefined. In this case, we need to find where the derivative f'(x) = 0 or f'(x) is undefined.
Step 2: Calculate the derivative, f'(x), of the function. Applying the quotient rule, the derivative is given by f'(x) = ((2(x - 2) - (2x - 3)(1))/((x - 2)^2).
Step 3: Set the derivative f'(x) equal to zero and solve for x, if possible. In this case, since the derivative is a rational function, we need to find where the numerator equals zero: 2(x - 2) - (2x - 3)(1) = 0.
Step 4: Determine the intervals using the critical points and the values around them. You can create a number line and test intervals on either side of the critical value(s), which are the values of x obtained in Step 3.
When you perform these steps for the function f(x) = 2x - 3/(x - 2), you would find the critical point at x = 2.
You can create a number line and test intervals to the left and right of x = 2. For example, if you test an interval to the left of x = 2, such as x = 1, you can plug it into f'(x) to determine if the derivative is positive or negative. Repeat this for intervals to the right as well.
When you perform these tests, you will find that the function f(x) = 2x - 3/(x - 2) is increasing to the left of x = 2 and decreasing to the right of x = 2. Therefore, the intervals on which the function is increasing are (-∞, 2) and the intervals on which the function is decreasing are (2, ∞).
2. To find the horizontal asymptotes of the function f(x) = 10/(x + 3), you can follow these steps:
Step 1: Determine the behavior of the function as x approaches positive and negative infinity. This will help identify if there are any horizontal asymptotes.
As x approaches positive infinity (x → +∞), the value of the function 10/(x + 3) will approach zero. This is because as the denominator gets larger (as x gets larger), the overall value of the function becomes smaller.
As x approaches negative infinity (x → -∞), the value of the function 10/(x + 3) will also approach zero. This is because as the denominator becomes more negative (as x gets more negative), the overall value of the function becomes smaller.
Step 2: Since the function approaches zero as x approaches positive and negative infinity, we can conclude that the function has a horizontal asymptote at y = 0.
Therefore, the horizontal asymptote of the function f(x) = 10/(x + 3) is y = 0.