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Differential Calculus> Q. GIVEN f( x ) = e x .cos x , x- [0,2pi...

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Q. GIVEN f( x ) = e^x.cos x , x- [0,2pi]. THE SLOPE OF TANGENT OF THE FUNCTION IS MINIMUM FOR-

x = pi

x = pi/4

x = 3pi/4

x = 3pi/2

Bharat Makkar , 11 Years ago

Grade Select Grade

2 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

$f(x) = e^xcosx$
$f'(x) = e^xcosx+e^x(-sinx)$
$f'(x) = e^x(cosx-sinx)$
We need to minimize this,
$f''(x) = e^x(cosx-sinx)+e^x(-sinx-cosx)$
$f''(x) = -2e^xsinx$
$2e^xsinx = 0$
$x =0, \pi , 2\pi$
So
$x = \pi$

Last Activity: 11 Years ago

Vicky

Pi/4 Will be the answer

Because sinx and cosine x will give minimum value at x=pi/4

So it will be OK for others

Last Activity: 7 Years ago

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