Find an equation for the tangent line to the
graph of f at the point P = (−1, f(−1)) when
f(x) =
4x + 1
3x + 1
What is f(x) ? I don't understand why you have two functions on top of one another. Both are straignt lines. Do you mean
f(x) = (4x + 1)/(3x + 1) ?
To find the equation of the tangent line to the graph of f at the point P = (-1, f(-1)), we need to find the derivative of f and evaluate it at x = -1.
First, let's find the derivative of f(x). The derivative of f(x) can be found using the quotient rule. The quotient rule states that for a function u(x)/v(x), the derivative is given by (v(x)*u'(x) - u(x)*v'(x))/(v(x))^2.
Applying the quotient rule to f(x) = (4x + 1)/(3x + 1), we have:
f'(x) = [(3x + 1)*(4) - (4x + 1)*(3)] / (3x + 1)^2
Now, let's evaluate f'(-1) to find the slope of the tangent line at x = -1.
f'(-1) = [(3*(-1) + 1)*(4) - (4*(-1) + 1)*(3)] / (3*(-1) + 1)^2
Simplifying the expression, we get:
f'(-1) = [(-2)*(4) - (-3)*(3)] / ((-2)^2)
f'(-1) = (-8 + 9) / 4
f'(-1) = 1/4
So, the slope of the tangent line at the point P is 1/4.
The equation of a line can be written in point-slope form:
y - y₁ = m(x - x₁)
Where (x₁, y₁) is the point on the line and m is the slope.
Substituting x₁ = -1, y₁ = f(-1), and m = 1/4, we have:
y - f(-1) = (1/4)(x - (-1))
y - f(-1) = 1/4(x + 1)
This is the equation of the tangent line to the graph of f at the point P = (-1, f(-1)).