Question

Let f(x)=2/1+x^2. The equation of the tangent line to the curve at the point (5,0.07692) can be written in the form y=mx+b

whats m and b?

Answer 1

Do you mean f(x)=2/(1+x^2) or (2/1)+x^2?

Answer 2

first one sorry

Answer 3

Assuming it's the former, taking the derivative of it would give us -4x(1+x^2)^-2. Plug in the point of x we're looking for to obtain the slope y'.

Answer 4

Is there a step you don't understand?

Answer 5

so i plug in 5 into the derivative?

Answer 6

Yes.

Answer 7

You use the original y(x) value to evaluate the term b in y=y'(5)x+b

Answer 8

can you explain that clearly please

Answer 9

Sure. We're given the original function y(x)=2/(1+x^2), right? We also know that y(5)=0.07692.

Now, we want to find the tangent line at y(5). The line will be given in the form y=mx+b. m is the slope, which we can find at any given point by the first derivative.

So, y'(x)=(-4x)/(1+x^2)^2. At y'(5), we can find our slope at y(5), which should be our m. So, now we know that y=y'(5)x+b, so now we need to solve for b. We know the original (x,y) for y(5), so we plug them in and voila, we have the values of b (in addition to m).