What are the critical values of #f(x)=3x^2+2x+5#?

Answer 1

#x=-1/3,y=14/3#

Computing the first derivative of #f(x)# we get
#f'(x)=6x+2# solving the equation #f'(x)=0# we get #x=-1/3#
since #f''(x)=6>0# we get a minimum (this is global, since we have a parabola) and we get
#f(-1/3)=3*1/9-2/3+5=(1-2+15)/3=14/3#
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Answer 2

To find the critical values of ( f(x) = 3x^2 + 2x + 5 ), we need to find the values of ( x ) where the derivative of ( f(x) ) is equal to zero or undefined.

  1. Find the derivative of ( f(x) ) with respect to ( x ):

[ f'(x) = 6x + 2 ]

  1. Set ( f'(x) ) equal to zero and solve for ( x ):

[ 6x + 2 = 0 ] [ 6x = -2 ] [ x = -\frac{2}{6} ] [ x = -\frac{1}{3} ]

  1. ( f'(x) ) is defined for all real numbers, so there are no critical values arising from points where the derivative is undefined.

Therefore, the only critical value of ( f(x) = 3x^2 + 2x + 5 ) is ( x = -\frac{1}{3} ).

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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.

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.

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.

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.

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