write the equation of the line tangent to the graph of f(x)= 1/(x^2 +2) at x=0
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Question

kainui · Answer

1/(x^2+2) = (x^2+2)^-1
That little shift changes it from having to apply the quotient rule to just having to use a simple chain rule. So let's go to the next step.

(-1(x^2+2)^-2)*(2x)
There's the derivative of the line, so plug in 0 to find the slope of the line at that point. It ends up having a slope of 0.

Any line with a slope of 0 is a straight line, so now we go back to the original function and plug in 0 to find the y-intercept (AKA 'B' of Y=MX+B)

We find that the equation for the tangent line should then be Y=0X+.5