Find the slope of the tangent line and the equation of the tangent line to f(x)=x^2-x at x=3
The slope of the tangent for any x is
f'(x) = 2x-1
So, you want the line
y-f(3) = f'(3) * (x-3)
We weren't allowed to just use derivatives for it we had to do this using limits. That's why I was unsure of this.
I guarantee that your text has this very problem in it. It starts out as the limit of
√(x+h)-√x
------------
h
That's hard to handle, but if you multiply top and bottom by √(x+h)+√x things will drop out nicely when you take the limit.
To find the slope of the tangent line and the equation of the tangent line at a specific point on a function, we need to take the derivative of the function.
Step 1: Find the derivative of the function f(x) = x^2 - x. The derivative gives us the slope of the tangent line at any point on the function.
f'(x) = 2x - 1
Step 2: Substitute the x-value of the point you're interested in, which is x = 3, into the derivative.
f'(3) = 2(3) - 1
= 6 - 1
= 5
So the slope of the tangent line at x = 3 is 5.
Step 3: Now that we have the slope, we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.
Plugging in the point (3, f(3)) = (3, 3^2 - 3) = (3, 6) and the slope m = 5 into the point-slope form, we get:
y - 6 = 5(x - 3)
Now we can simplify this equation to find the equation of the tangent line:
y - 6 = 5x - 15
y = 5x - 9
Therefore, the equation of the tangent line to f(x) = x^2 - x at x = 3 is y = 5x - 9.