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What is the instantaneous rate of change of #f(x)=sqrt(3x^2+4x+2) # at #x=4 #?

Answer 1

#f'(4)=28/(2sqrt66)=1.72328#.

The instantaneous rate of change at a point is equivalent to the derivative evaluated at that point. We evaluate this derivative by using the power rule.

#therefore f'(x)=d/dx(3x^2+4x+2)^(1/2)#

#=1/2(3x^2+4x+2)^(-1/2)*(6x+4)#

#therefore f'(4)=(6(4)+4)/(2sqrt((3(4^2)+4(4)+2)))=28/(2sqrt66)=1.72328#.

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Answer 2

Samantha Adkison

To find the instantaneous rate of change of ( f(x) = \sqrt{3x^2 + 4x + 2} ) at ( x = 4 ), we first find the derivative of ( f(x) ) with respect to ( x ), which is ( f'(x) = \frac{6x + 4}{2\sqrt{3x^2 + 4x + 2}} ). Then, we evaluate ( f'(4) ) to find the instantaneous rate of change at ( x = 4 ). Plugging in ( x = 4 ) into the derivative, we get ( f'(4) = \frac{6(4) + 4}{2\sqrt{3(4)^2 + 4(4) + 2}} = \frac{24 + 4}{2\sqrt{48 + 16 + 2}} = \frac{28}{2\sqrt{66}} = \frac{14}{\sqrt{66}} ). So, the instantaneous rate of change of ( f(x) ) at ( x = 4 ) is ( \frac{14}{\sqrt{66}} ).

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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.