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Mathematics 13 Online
OpenStudy (anonymous):

If y=|cos x| then f'(3pi/4) is?

OpenStudy (anonymous):

\[\frac{-1}{\sqrt{2}}\]

OpenStudy (anonymous):

it should be +ve 1/root2

OpenStudy (anonymous):

no, i don't think

OpenStudy (anonymous):

the slope of the tangent is negative at the reqd point in the graph of f(x)=|cosx|

OpenStudy (anonymous):

sorry, its 1/rt(2). u r right.

OpenStudy (anonymous):

ok how did you get it

OpenStudy (anonymous):

in the interval (pi/2,pi) the function is f(x)=+cosx so its derivative is sinx and at 3pi/4 its value is 1/rt(2). Observe that i had to find the sign of the function in the neighbourhood of the given point.

OpenStudy (anonymous):

f(x)=-cos x in the intervl pi/2 to pi

OpenStudy (anonymous):

yes... so the value is positive.

OpenStudy (anonymous):

i mean when u plug pi/2 in f(x), f(x) comes positive

OpenStudy (anonymous):

f(x)=-cosx in pi/2 to pi derivative of cosx is -sinx so f'(x)=sinx

OpenStudy (anonymous):

ok..got it

OpenStudy (anonymous):

bingo

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