How do you find the average rate of change of #y=x^3+1# from #x=1# to #x=3#?

Answer 1
#(f(x+h)-f(x))/h=(f(b)-f(a))/(b-a)#, where #a# is the lower bound and #b# is the upper bound.

The mean rate of variation

#slope=(f(b)-f(a))/(b-a)=(f(3)-f(1))/(3-1)=((3)^3+1-((1)^3+1))/(3-1)=((27+1)-(1+1))/(3-1)=(28-2)/(3-1)=26/2=13#
Point: #(3,28)# Point: #(1,2)#
#y=mx+b#
#2=13(1)+b#
#2=13+b#
#-11=b#
#y=13x-11#, the secant line through the points #(3,28)# and #(1,2)#.
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Answer 2

To find the average rate of change of ( y = x^3 + 1 ) from ( x = 1 ) to ( x = 3 ), first find the values of ( y ) at ( x = 1 ) and ( x = 3 ).

  1. Substitute ( x = 1 ) into the equation: ( y(1) = (1)^3 + 1 = 1 + 1 = 2 ).

  2. Substitute ( x = 3 ) into the equation: ( y(3) = (3)^3 + 1 = 27 + 1 = 28 ).

Then, calculate the change in ( y ) over the interval ( x = 1 ) to ( x = 3 ), which is ( \Delta y = y(3) - y(1) = 28 - 2 = 26 ).

The change in ( x ) over the same interval is ( \Delta x = 3 - 1 = 2 ).

Finally, divide the change in ( y ) by the change in ( x ) to find the average rate of change: [ \text{Average rate of change} = \frac{\Delta y}{\Delta x} = \frac{26}{2} = 13. ]

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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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