# calculus

posted by on .

To find the equation of a line, we need the slope of the line and a point on the line.

Since we are requested to find the equation of the tangent line at the point (64, 8), we know that (64, 8) is a point on the line. So we just need to find its slope.

The slope of a tangent line to f(x) at x = a can be found using the formula
mtan = lim x→a f(x) − f(a)/x − a

In this situation, the function is
f(x) = a=

• calculus - eh? - ,

so far, so good. What's the question?

• calculus - ,

its asking for f(x) and a

im just trying to figure out how i would go about doing this. thanks for answering me.

• calculus - ,

i got a slope of 1/16 is that right?

• calculus - ,

since I have no idea what f(x) is, I can't say. It never appears in the exercise posted.

• calculus - ,

maybe if you showed your calculation of the slope I can figure out how you got it, and whether it's right.

• calculus - ,

well the question is from what im assuming is trying to find what f(x) is equal to. that's the last line of my question above. it also wants me to find what a is equal to. so to find slope i took the first derivative: y'= 1/ [2sqrt(x)] since x=64
makes this y'=1/(2*8) = 1/16

• calculus - ,

this sentence doesnt look quite right it should look more like this

In this situation, the function is
f(x) = ? a= ?