# Thread: help with calculus problem: Finding equation of tangent line at given point

1. ## help with calculus problem: Finding equation of tangent line at given point

Hi, I'm new to this forum and I need some help with this problem. If you could help, I'd appreciate it.

Find an equation of the tangent line at the point indicated.
f (x) = log2 (3x + x^-3), at x = 1

I tried using y - y1 = m(x - x1)

for y1, i got 2, x1 is 1, and for the slope i got (3-3/x^4)/((1/x^3+3 x) log(2))

so for the equation of the line, I got ((3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)) +2

but unfortunately, its wrong.. lol

2. Originally Posted by justin144
Hi, I'm new to this forum and I need some help with this problem. If you could help, I'd appreciate it.

Find an equation of the tangent line at the point indicated.
f (x) = log2 (3x + x^-3), at x = 1

I tried using y - y1 = m(x - x1)

for y1, i got 2, x1 is 1, and for the slope i got (3-3/x^4)/((1/x^3+3 x) log(2))

so for the equation of the line, I got ((3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)) +2

but unfortunately, its wrong.. lol
m = f'(1) = ....

f(1) = 2 so use x1 = 1 and y1 = 2.

3. right. this is what I did. so I got :
y-2 = (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)
add two to both sides which gives:
y= (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1) +2
but the program is still counting my answer as incorrect.
So did I get the slope wrong?

4. Originally Posted by justin144
right. this is what I did. so I got :
y-2 = (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)
add two to both sides which gives:
y= (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1) +2
but the program is still counting my answer as incorrect.
So did I get the slope wrong?
For the second time:

m = f'(1) = ....

Do you know what this means? It means the value of the derivative at x = 1.

5. Originally Posted by mr fantastic
For the second time:

m = f'(1) = ....

Do you know what this means? It means the value of the derivative at x = 1.
ohhh ok. so plugging x=1 into the derivative, i get:
m=0, sooo:
y-2 = 0(x-1)
which is: y=2
Ok, i'll try that. Thank you.