Question

differentiate the functions n find the slope of the tang. line at the given value, f(x)=x+(9/x), x=-3, how would i set this up? f(x+h)-f(x)/h for the slope? what about the first part? help!!

Answer 1

no, you're wrong....log(x) is not the derivative of f(x)....

Answer 2

well chemmaj88, if you want to find the slope of the tangent line, you have to find the first derivative of this function, and then plug in x=-3....do you know how to do that?

Answer 3

Take the derivative of the function f(x)=f'(x) (and derive it correctly). Evaluate f'(x) at x=a,f'(a), to obtain the slope at x=a. Get an original pair of coordinates for f(x) (x,f(x)) and plug this in for y=f'(a)x+y_0 to solve for y_0, or the y-intercept.

Answer 4

That's the condensed instructions, other people might be more helpful. ;D

Answer 5

so thats why i writen first derivative of f(x) & then putt the value x=-3

Answer 6

\[f \prime(x)=1-9/x^2\] so \[f \prime(-3)=1-9/9=0\] so the slope of the tangent line is 0 so that means that the tangent line is horizontal

Answer 7

how did you get the 1-9/x^2?

Answer 8

i dont knw how to get the deriv.

Answer 9

now, if you want to find the equation of the tangent line, besides the slope you're going to need the point...plug in x=-3 in f(x) and you'll get -3+9/(-3)=-3-3=-6 so the equation is: \[y+6=0*(x+3)\] and y=-6 well you must know the rules for basic derivatives...:D