Asked by Anonymous
Find an equation of the tangent line to the graph of f(x)=x+3/x^2+2
at x=1
A.5x+9y=17
B.5x+3y=9
C.9y-5x=7
D.5x+9y=7
at x=1
A.5x+9y=17
B.5x+3y=9
C.9y-5x=7
D.5x+9y=7
Answers
Answered by
Reiny
I will assume you meant:
f(x)= (x+3)/(x^2+2)
then f(1) = 4/3, so the point you want is (1, 4/3)
f'(x) = ( (x^2 + 2)(1) - (x+3)(2x) )/(x^2 + 2)^2
f'(1) = (3 - 8)/9 = - 5/9
tangent: y - 4/3 = (-5/9)(x - 1)
simplify to see which of the choices matches
f(x)= (x+3)/(x^2+2)
then f(1) = 4/3, so the point you want is (1, 4/3)
f'(x) = ( (x^2 + 2)(1) - (x+3)(2x) )/(x^2 + 2)^2
f'(1) = (3 - 8)/9 = - 5/9
tangent: y - 4/3 = (-5/9)(x - 1)
simplify to see which of the choices matches
Answered by
Sarah
Find an equation of the line tangent to the graph of x^2+(y-x)^3=9 at x=1
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