1. find the equation of the tangent line at x=1 to y=f(x) where f(x)=(3x^2)/(5x^2+7x)
Can someone help walk me through the steps of this problem. Do I start with finding the derivative of the function using the quotient rule?
Yes, you are correct. To find the equation of the tangent line at x=1 to the function y=f(x), you need to find the derivative of the function using the quotient rule.
Let's start by finding the derivative of f(x). The quotient rule states that if you have a function f(x) = g(x)/h(x), then the derivative of f(x) is given by:
f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2
In our case, g(x) = 3x^2 and h(x) = 5x^2 + 7x. Let's differentiate g(x) and h(x) separately:
g'(x) = d/dx (3x^2) = 6x
h'(x) = d/dx (5x^2 + 7x) = 10x + 7
Now we can substitute these values into the quotient rule to find f'(x):
f'(x) = (6x * (5x^2 + 7x) - 3x^2 * (10x + 7)) / ((5x^2 + 7x)^2)
Simplifying this equation further:
f'(x) = (30x^3 + 42x^2 - 30x^3 - 21x^2) / ((5x^2 + 7x)^2)
f'(x) = (21x^2) / ((5x^2 + 7x)^2)
Now that we have the derivative f'(x), we can find the slope of the tangent line by plugging in the value x=1:
slope = f'(1) = (21(1)^2) / ((5(1)^2 + 7(1))^2)
slope = 21 / (5 + 7)^2 = 21/12^2 = 21/144
Therefore, the slope of the tangent line at x=1 is 21/144.
To find the equation of the tangent line, we need both the slope and a point on the line. We already know the x-coordinate of the point, which is x=1. To find the corresponding y-coordinate, we can substitute x=1 back into the original function f(x):
f(1) = (3(1)^2) / (5(1)^2 + 7(1)) = 3/12 = 1/4
Therefore, the y-coordinate of the point is y=1/4.
Now we can use the point-slope form of a line to find the equation:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is the point on the line.
Plugging in the values we have:
y - 1/4 = (21/144)(x - 1)
Multiplying both sides by 144 to eliminate the fraction:
144y - 36 = 21(x - 1)
Rearranging the equation:
21x - 144y = 165
This is the equation of the tangent line at x=1 to the function y=f(x).