find the eq to the tangent line to the curve y=x^3 g(x) at x=1 if g(1)=2 f(1)=3 f'(1)=1 g'(1)=-2
i cant understand how the answer is y=4x-2
To find the equation of the tangent line to the curve y = g(x) at x = 1, we need to determine the slope of the tangent line and a point on the line.
The slope of the tangent line can be found by evaluating the derivative of g(x) at x = 1. In this case, g'(1) = -2, which represents the slope of the tangent line.
To find a point on the line, we can substitute the coordinates (1, g(1)) into the equation y = mx + b, where m is the slope and b is the y-intercept.
Given that g(1) = 2, we substitute these values into the equation:
y = -2x + b
Using the second set of information, we can find b.
Given that f(1) = 3, we can also find the slope of the tangent line to f(x) at x = 1. The slope is represented by f'(1), which is equal to 1.
Using the point-slope form of a line, we can substitute the coordinates (1, f(1)) and the slope f'(1) = 1 into the equation:
y - f(1) = f'(1)(x - 1)
y - 3 = 1(x - 1)
y - 3 = x - 1
y = x + 2
Now we have two equations:
(1) y = -2x + b
(2) y = x + 2
To find the value of b, we can equate the two equations since they represent the same tangent line at the same point (x = 1).
-x + 2 = -2x + b
-2 = -x + b
To find b, we can substitute the value of x with 1 and solve for b:
-2 = -1 + b
b = -1 + (-2)
b = -3
Now we have determined the value of b, and we can rewrite the equation (1) as:
y = -2x - 3
Hence, the equation of the tangent line to the curve y = g(x) at x = 1 is y = -2x - 3.