find the equation of tangent line to the graph of f(x)=xcosx(√2 x) at a=0
6 answers
sorry the question is f(x)=xcos(√2 x) at a=0
well,
f'(x) = cos(√2 x) - √2 x sin(√2 x)
Not sure what a has to do with things, but if you meant x=0, then
f(0) = 0
f'(0) = 1
So, you want the line with slope 1, which passes through (0,0).
No too much of a challenge there, eh?
f'(x) = cos(√2 x) - √2 x sin(√2 x)
Not sure what a has to do with things, but if you meant x=0, then
f(0) = 0
f'(0) = 1
So, you want the line with slope 1, which passes through (0,0).
No too much of a challenge there, eh?
yep thanks
it will be y=x but how do we know it passes through (0,0)
Because y(0) = 0. It helps if you actually read what I write. Or maybe it has something to do with the mysterious "a".
okay I get it and a is as you said just to say x