I'm feeling pretty stupid right now because I can't remember how to do these.
if f(x) = (2x-5)/(x^2-4), write an equation of the tangent line to f at point x=0. *How do I find the slope of the tangent line?
and how do you put x=ln4 into a y= equation?
Take the derivative of f(x), that is the slope.
The second question if y is a function of x, then put in ln4 for the value of x to be determined.
Finding the slope of the tangent line at a specific point and expressing an equation in the form y = f(x) are common tasks in calculus. Let me explain step-by-step how to answer these questions:
1. Finding the slope of the tangent line:
To find the slope of the tangent line, you need to take the derivative of the function f(x). In this case, f(x) = (2x - 5) / (x^2 - 4). Taking the derivative of this function will give you the slope of the tangent line at any given point x.
To find the derivative, you can use the quotient rule. Let me break it down for you:
- Differentiate the numerator: (d/dx)(2x - 5) = 2
- Differentiate the denominator: (d/dx)(x^2 - 4) = 2x
- Apply the quotient rule: (2(x^2 - 4) - (2x - 5)(2x)) / (x^2 - 4)^2
Now, to find the slope at a specific point, substitute x = 0 into the derivative expression:
slope = (2(0^2 - 4) - (2(0) - 5)(2(0))) / (0^2 - 4)^2
Simplifying this expression will give you the slope of the tangent line at x = 0.
2. Substituting x = ln(4) into a y = equation:
When you are given a value for x and need to find the corresponding y-value in a given equation, you can substitute the value into the equation and solve for y.
In this case, you have x = ln(4), and you want to find the corresponding y-value in a y = equation. Just replace x with ln(4) in the equation and solve for y.
For example, if you have the equation y = 2x^2 + 3x + 1, you can substitute x = ln(4) into the equation:
y = 2(ln(4))^2 + 3(ln(4)) + 1
Evaluating this expression will give you the corresponding y-value when x = ln(4).
I hope these explanations help! Let me know if you have any further questions.