using first principle, find the slope of a tangent line to the curve f(x) = x^2-3x+5 at x=3
find the limit of
(f(x+h)-f(x))/h
= [((x+h)^2 - 3(x+h) + 5) - (x^2-3x+5)]/h
You should wind up with 2x-3
To find the slope of the tangent line to the curve at a specific point using the first principle, we can follow these steps:
Step 1: Write down the given function:
f(x) = x^2 - 3x + 5
Step 2: Differentiate the function with respect to x:
f'(x) = (d/dx) (x^2 - 3x + 5)
Step 3: Apply the power rule of differentiation where d/dx(x^n) = n*x^(n-1):
f'(x) = 2x - 3
Step 4: Plug the value of x into the derivative to find the slope at that point:
Slope at x=3: f'(3) = 2(3) - 3 = 6 - 3 = 3
Therefore, the slope of a tangent line to the curve f(x) = x^2 - 3x + 5 at x=3 is 3.
To find the slope of the tangent line to a curve at a given point using first principles, we need to follow these steps:
1. Determine the equation of the curve: The given curve is f(x) = x^2 - 3x + 5.
2. Find the derivative of the curve: The derivative gives the slope of the tangent line at any point on the curve.
To find the derivative of f(x), we can apply the power rule for differentiation. According to the power rule, the derivative of x^n with respect to x is n*x^(n-1). Since f(x) = x^2 - 3x + 5, we can differentiate term by term:
f'(x) = d/dx (x^2 - 3x + 5)
= d/dx (x^2) - d/dx (3x) + d/dx (5)
= 2x - 3 + 0
= 2x - 3
3. Evaluate the derivative at x = 3: Substitute x = 3 into the derivative function f'(x) to find the slope at that specific point.
f'(3) = 2(3) - 3
= 6 - 3
= 3
So, the slope of the tangent line to the curve f(x) = x^2 - 3x + 5 at x = 3 is 3.