tangent line ARG

notamathperson

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May 25, 2014
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Hello could you help me solve this : f(x)=e^(x^2)cos (x) I have to find the equation of the tangent line at x=0 I understand that I have to find the derivative and plug in zero, but after I am stuck :(:(
 
Any (non-vertical) line can be written in the form y= mx+ b. Finding the derivative and evaluating at x= 0 will give you the slope, m. To find b note that at x= 0, \(\displaystyle y= e^0cos(0)= 1\). Putting x= 0, y= 1 in y= mx+ b tells you that b= 1.
 
Any (non-vertical) line can be written in the form y= mx+ b. Finding the derivative and evaluating at x= 0 will give you the slope, m. To find b note that at x= 0, \(\displaystyle y= e^0cos(0)= 1\). Putting x= 0, y= 1 in y= mx+ b tells you that b= 1.
How come when you plug in 0 it gives you 1 :confused:
 
How come when you plug in 0 it gives you 1

In the first case, f(0) = 1 because f(0) is defined as e^0*cos(0)

e^0 and cos(0) are both equal to 1, so f(0) is 1*1

In the second case, b=1 because substituting x=0 and y=1 into the equation y=m*x+b gives 1=0+b which simplifies to b=1.

:smile:
 
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