How do I do this question: find slope of the curve y=1/(X-1) at X=2. I don't get the algebra here, how is it equal to -1?
Thanks
Oh and please do not answer with derivatives. Need to use the (f(x-h)-f(x))/h
well, just plug and chug.
f(x+h)-f(x) = 1/(x+h-1) - 1/(x-1)
= ((x-1)-(x+h-1))/((x-1)(x+h-1))
= (x-1-x-h+1)/((x-1)(x+h-1))
= -h/((x-1)(x+h-1))
Now, divide that by h, and you have
-1/((x-1)(x+h-1))
Now you can take the limit as h->0, and you end up with
-1/(x-1)^2
at x=2, that is -1
To find the slope of the curve y=1/(x-1) at x=2, we'll use the concept of derivatives. The slope of a curve at a specific point is defined as the slope of the tangent line to the curve at that point.
To start, let’s find the derivative of the given curve. The derivative of y with respect to x, denoted as dy/dx or y', represents the rate of change of y with respect to x.
To find the derivative, we can use the power rule for differentiation. For a function of the form f(x) = 1/x, the derivative is given by:
f'(x) = -1/x^2
Applying this rule to the given equation y = 1/(x-1), we get:
dy/dx = -1/(x-1)^2
Now, to find the slope at a specific point, we substitute the given value of x into the derivative expression.
Let's evaluate the derivative at x = 2:
dy/dx = -1/(2-1)^2 = -1/1^2 = -1/1 = -1
Therefore, the slope of the curve y=1/(x-1) at x=2 is -1.
In summary, to find the slope of a curve at a specific point, you need to differentiate the equation of the curve with respect to x, and then substitute the given value of x into the derivative expression.