anonymous
  • anonymous
Suppose that f has a continuous second derivative for all x, and that f(0) = 1, f ' (0) = 2, and f''(0) = 0. A. Does f have an inflection point at x = 0? Explain your answer. B. Let g'(x) = (3x^2 + 2)f(x) + (x^3 + 2x + 5)f'(x). The point (0,5) is on the graph of g. Write the equation of the tangent line to g at this point. C. Use your tangent line to approximate g(0. 3). D. Find g''(0).
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
I feel like A is no because f'(0) would have to = 0 for it to be a critical point? Is that correct?
dan815
  • dan815
no it can be an inflection point even if its not a critical point
dan815
  • dan815
|dw:1435145399302:dw|

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dan815
  • dan815
inflection points are where the concavity flips, and critical point is where slope is 0
dan815
  • dan815
and roots are where the function is 0
Michele_Laino
  • Michele_Laino
I thin|dw:1435145496095:dw|k that x=0 is an inflection point with inclined tangent, like this:
anonymous
  • anonymous
And concavity is usually determined by the second derivative? So what does it mean that this second derivative is always 0?
Michele_Laino
  • Michele_Laino
oops.. ...I think that x=0 is an inflection point with inclined tangent, like this...
Michele_Laino
  • Michele_Laino
the second derivative is zero only at x=0 so we can not say that there exist a complete neighborhood in which f ''(x) >0 and we can not say that there exists a complete neighborhood in which f'' (x) <0
anonymous
  • anonymous
Oh, I totally misread the question! sorry
anonymous
  • anonymous
So how should I be testing for it being an inflection point?
Michele_Laino
  • Michele_Laino
since f ' (2) is positive, then we can find a complete neighborhood of the point x=2 such that the function f is an increasing function
Michele_Laino
  • Michele_Laino
so, being f ''(2)=0 in that complete neighborhood there are points for which f ''(x) >0 and points for which f ''(x) <0
anonymous
  • anonymous
Which would make it an inflection point because there are values in either direction at f''(x)? Or am I jumping ahead here?
Michele_Laino
  • Michele_Laino
|dw:1435146282570:dw|
anonymous
  • anonymous
I see! That makes sense
Michele_Laino
  • Michele_Laino
you are right, we have an inflection point since there are values in either direction of f''(x), inside a complete neighborhood of x=2
anonymous
  • anonymous
Great. So the derivative of g(x) is really throwing me off in the next part. But I just plug in f(0) and f'(0) and (0,5), right? then use that as a slope for a y-y1=m(x-x1) formula for the tangent line?
Michele_Laino
  • Michele_Laino
yes! that's right you should get this:
Michele_Laino
  • Michele_Laino
\[y - 5 = 2\left( {x - 5} \right)\]
anonymous
  • anonymous
great! So then to find g(.3) using a tangent line approximation, do i plug in .3=x to that tangent line equation?
Michele_Laino
  • Michele_Laino
oops...wrong equation. Here is the right equation \[y - 5 = 2x\]
Michele_Laino
  • Michele_Laino
we can write this: \[g\left( {0.3} \right) - g\left( 0 \right) \cong g'\left( 0 \right)\left( {0.3 - 0} \right)\]
anonymous
  • anonymous
where does the g(0) come from?
Michele_Laino
  • Michele_Laino
since by thefinition of first derivative we can write: \[g\left( x \right) - g\left( {{x_0}} \right) = g'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + o\left( {x - {x_0}} \right)\]
Michele_Laino
  • Michele_Laino
where: \[o\left( {x - {x_0}} \right)\] is a quantity which goes to zero as x approaches to x_0
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
so substituting your data we can write: \[g\left( {0.3} \right) = 5 + 0.6 = 5.6\]
anonymous
  • anonymous
Awesome! And then for g''(x)... does this look correct: 6x^2+6x+4 so that g''(0)=4
Michele_Laino
  • Michele_Laino
Starting for m your formula for g '(x), after simple computation, I got this:
Michele_Laino
  • Michele_Laino
\[g''\left( x \right) = 6xf\left( x \right) + \left( {6{x^2} + 4} \right)f'\left( x \right) + \left( {{x^3} + 2x + 5} \right)f''\left( x \right)\]
Michele_Laino
  • Michele_Laino
so, substituting x=0, we can write: \[g''\left( 0 \right) = 4\]
Michele_Laino
  • Michele_Laino
oops.. starting from your formula...
anonymous
  • anonymous
Thank you so much! This has been a great help!!
Michele_Laino
  • Michele_Laino
:)
Michele_Laino
  • Michele_Laino
please wait, I have made an error, we have: g' (0) = 2+10=12 so the equation for tangent line is: \[y - 5 = 12x\]
Michele_Laino
  • Michele_Laino
sorry for my error of computation
Michele_Laino
  • Michele_Laino
I have to be grateful to @Loser66 , who kindly has warned me. Thanks !! @Loser66
anonymous
  • anonymous
so g(.3) now is 8.6, yes?
Michele_Laino
  • Michele_Laino
yes! we have: \[g\left( {0.3} \right) = 5 + 12 \times 0.3 = 8.6\]
anonymous
  • anonymous
Thank you!!
Michele_Laino
  • Michele_Laino
:)

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