Asked by selina
                find the equation of tangent line to the graph of f(x)=xcosx(√2 x) at a=0
            
            
        Answers
                    Answered by
            selina
            
    sorry the question is f(x)=xcos(√2 x) at a=0
    
                    Answered by
            Steve
            
    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?
                    Answered by
            selina
            
    yep thanks 
    
                    Answered by
            selina
            
    it will be y=x but how do we know it passes through (0,0)
    
                    Answered by
            Steve
            
    Because y(0) = 0. It helps if you actually read what I write. Or maybe it has something to do with the mysterious "a".
    
                    Answered by
            selina
            
    okay I get it and a is as you said just to say x
    
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