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How do you find critical points for the equation #(7/10)cos(x/10)#?

Answer 1

Charles Anctil

#x=10kpi# where #k# is an integer.

#f(x)=7/10cos(x/10)#

Critical points are when derivative is 0 or undefined.

#f'(x)=-7/100sin(x/10)#

Sin is well defined at every point but is equal to zero at 0,#pi#

#-7/100sin(x/10)=0#

#sin(x/10)=0

#sin(y)=0 # at #y=kpi#

Therefore

#sin(x/10)=0# at # x/10=kpi# so #x=10kpi# where #k# in an integer.

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Answer 2

Paisley Johnson

To find the critical points of the function ( f(x) = \frac{7}{10} \cos\left(\frac{x}{10}\right) ), we first need to find the derivative of the function, then set it equal to zero and solve for ( x ).

The derivative of ( f(x) ) with respect to ( x ) is ( f'(x) = -\frac{7}{100} \sin\left(\frac{x}{10}\right) ).

To find critical points, we set ( f'(x) = 0 ) and solve for ( x ): [ -\frac{7}{100} \sin\left(\frac{x}{10}\right) = 0 ]

Since sine function equals zero at multiples of ( \pi ), we have: [ \frac{x}{10} = k\pi ] Where ( k ) is an integer.

Solving for ( x ), we get: [ x = 10k\pi ]

Therefore, the critical points occur at ( x = 10k\pi ), where ( k ) is an integer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.