y–y0=f'(x0)*(x–x0)
f'=1/(3x+1)^2
x0=2
y0=2/(3*2+1)=2/(6+1)=2/7
f'(x0)=1/(3*2+1)^2=1/(6+1)^2=1/7^2=1/49
y–y0=f'(x0)*(x–x0)
y - 2/7 = 1/49*(x-2)
y=(1/49)*x - 2/49 + 2/7
y=(1/49)*x - 2/49 + 14/49
y=(1/49)*x - 12/49
y=(1/49)*(x-12)
For the function f(x) = x/(3x+1) find the equation of the line which is tangent to the curve f(x) at x=2. Solve analytically (an exact answer is required)
Please show steps.
Thanks.
2 answers
If You don't know how find derivation of x/(3x+1) in google type: calc101
When you open page calc101com click option derivatives.In rectangle type:
x/(3x+1)
and click option DO IT
You will see solution step by step
When you open page calc101com click option derivatives.In rectangle type:
x/(3x+1)
and click option DO IT
You will see solution step by step