Write the function that has derivative y prime = #x^2# and passes through (1,1) ?

Answer 1

#implies y = 1/3 (x^3 + 2)#

I'd follow the instructions on the packet.

#y' = x^2#
#implies y = x^3/3 + C#

Applying the IV's:

#y(1) = 1 implies 1 = 1/3 + C implies C = 2/3#

So:

#implies y = x^3/3 + 2/3 = 1/3 (x^3 + 2)#
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Answer 2

#y=(x^3)/3+2/3#
#color(white)("XXXXXXXXX")#although, maybe, that's only my opinion.

If #y' = x^2# then #color(white)("XXX")y=(x^3)/3+c# for some constant #c#
If #(color(red)x,color(blue)y)=(color(red)1,color(blue)1)# is a solution to this equation #color(white)("XXX")color(blue)1=color(red)1^3/3+c=1/3+c#
#color(white)("XXXXXX")rarr c=2/3#
and the function is #color(white)("XXX")y=x^3/3+2/3#
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Answer 3

The function that satisfies the given conditions is y = (1/3)x^3 + (2/3)x + 2/3.

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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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