J jyoun125 New member Joined Nov 12, 2014 Messages 8 Dec 8, 2014 #1 Let f(x)=x3-2x+4. Find an equation for the line tangent to the graph of f at the point (1,f(1)).
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Dec 8, 2014 #2 jyoun125 said: Let f(x)=x3-2x+4. Find an equation for the line tangent to the graph of f at the point (1,f(1)). Click to expand... Please show us some of your effort to complete this question.
jyoun125 said: Let f(x)=x3-2x+4. Find an equation for the line tangent to the graph of f at the point (1,f(1)). Click to expand... Please show us some of your effort to complete this question.
J jyoun125 New member Joined Nov 12, 2014 Messages 8 Dec 8, 2014 #3 pka said: Please show us some of your effort to complete this question. Click to expand... I know that f'(x)=3x2-2 and that f'(1)=1.
pka said: Please show us some of your effort to complete this question. Click to expand... I know that f'(x)=3x2-2 and that f'(1)=1.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Dec 8, 2014 #4 jyoun125 said: I know that f'(x)=3x2-2 and that f'(1)=1. Click to expand... \(\displaystyle (y-f(1))=f'(1)(x-1)\)
jyoun125 said: I know that f'(x)=3x2-2 and that f'(1)=1. Click to expand... \(\displaystyle (y-f(1))=f'(1)(x-1)\)
J jyoun125 New member Joined Nov 12, 2014 Messages 8 Dec 8, 2014 #5 So, since f(1)=3 that means that the equation of the line is y-3=1(x-1) or y=4x-4.
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,250 Dec 9, 2014 #6 jyoun125 said: So, since f(1)=3 that means that the equation of the line is y-3=1(x-1) or y=4x-4. \(\displaystyle \ \ \ \ \) Did you do this in your head? Click to expand... y - 3 = 1(x - 1) y - 3 = x - 1 y - 3 + 3 = x - 1 + 3 y = x + 2
jyoun125 said: So, since f(1)=3 that means that the equation of the line is y-3=1(x-1) or y=4x-4. \(\displaystyle \ \ \ \ \) Did you do this in your head? Click to expand... y - 3 = 1(x - 1) y - 3 = x - 1 y - 3 + 3 = x - 1 + 3 y = x + 2
