Using the definition of the derivative, find the equation of the tangent line to y=2x-1/2 at (1/2,-1)
Thanks!!
The equation you gave is a straight line.
There is no such thing as a tangent to a straight line, since that would have to be the line itself.
Since the point given is not on your given line, I suspect a typo
did you mean
y = 2x^(-1/2) ?
To find the equation of the tangent line to the function y = 2x - 1/2 at the point (1/2, -1), we can use the definition of the derivative.
The derivative of a function represents the rate of change of the function at a given point. In this case, we need to find the derivative of the function y = 2x - 1/2.
The derivative, denoted as dy/dx or f'(x), gives us the slope of the tangent line at any point on the curve. To find the derivative of y = 2x - 1/2, we differentiate the function with respect to x.
d(y)/dx = d(2x - 1/2)/dx
To differentiate, we use the power rule, which states that the derivative of x^n is n * x^(n-1).
d(2x - 1/2)/dx = 2 * d(x)/dx - 1/2 * d(1)/dx
Since d(x)/dx = 1 and d(1)/dx = 0 (as the derivative of a constant is always zero), we can simplify:
2 * d(x)/dx - 1/2 * d(1)/dx = 2 - 0 = 2
Therefore, the derivative of y = 2x - 1/2 is 2. The value of the derivative represents the slope of the tangent line to the curve at any given point.
To determine the equation of the tangent line at the point (1/2, -1), we know the slope of the line is 2 and we have a point (1/2, -1). We can use the point-slope form of a line to find the equation:
y - y1 = m(x - x1)
where (x1, y1) represents the coordinates of a point on the line, and m represents the slope of the line.
Substituting the values into the equation:
y - (-1) = 2(x - 1/2)
Simplifying further:
y + 1 = 2x - 1
y = 2x - 2 - 1
y = 2x - 3
Hence, the equation of the tangent line to y = 2x - 1/2 at the point (1/2, -1) is y = 2x - 3.