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anonymous
 one year ago
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).
anonymous
 one year ago
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).

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I feel like A is no because f'(0) would have to = 0 for it to be a critical point? Is that correct?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1no it can be an inflection point even if its not a critical point

dan815
 one year ago
Best ResponseYou've already chosen the best response.1inflection points are where the concavity flips, and critical point is where slope is 0

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and roots are where the function is 0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I thindw:1435145496095:dwk that x=0 is an inflection point with inclined tangent, like this:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And concavity is usually determined by the second derivative? So what does it mean that this second derivative is always 0?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4oops.. ...I think that x=0 is an inflection point with inclined tangent, like this...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4the 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I totally misread the question! sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how should I be testing for it being an inflection point?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4since 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
 one year ago
Best ResponseYou've already chosen the best response.4so, being f ''(2)=0 in that complete neighborhood there are points for which f ''(x) >0 and points for which f ''(x) <0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which would make it an inflection point because there are values in either direction at f''(x)? Or am I jumping ahead here?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435146282570:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see! That makes sense

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4you 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
 one year ago
Best ResponseYou've already chosen the best response.0Great. 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 yy1=m(xx1) formula for the tangent line?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4yes! that's right you should get this:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4\[y  5 = 2\left( {x  5} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0great! So then to find g(.3) using a tangent line approximation, do i plug in .3=x to that tangent line equation?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4oops...wrong equation. Here is the right equation \[y  5 = 2x\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4we can write this: \[g\left( {0.3} \right)  g\left( 0 \right) \cong g'\left( 0 \right)\left( {0.3  0} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where does the g(0) come from?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4since 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
 one year ago
Best ResponseYou've already chosen the best response.4where: \[o\left( {x  {x_0}} \right)\] is a quantity which goes to zero as x approaches to x_0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4so substituting your data we can write: \[g\left( {0.3} \right) = 5 + 0.6 = 5.6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! And then for g''(x)... does this look correct: 6x^2+6x+4 so that g''(0)=4

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4Starting for m your formula for g '(x), after simple computation, I got this:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4\[g''\left( x \right) = 6xf\left( x \right) + \left( {6{x^2} + 4} \right)f'\left( x \right) + \left( {{x^3} + 2x + 5} \right)f''\left( x \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4so, substituting x=0, we can write: \[g''\left( 0 \right) = 4\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4oops.. starting from your formula...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! This has been a great help!!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4please 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
 one year ago
Best ResponseYou've already chosen the best response.4sorry for my error of computation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I have to be grateful to @Loser66 , who kindly has warned me. Thanks !! @Loser66

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so g(.3) now is 8.6, yes?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4yes! we have: \[g\left( {0.3} \right) = 5 + 12 \times 0.3 = 8.6\]
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