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

- schrodinger

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- 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

no it can be an inflection point even if its not a critical point

- dan815

|dw:1435145399302:dw|

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## More answers

- dan815

inflection points are where the concavity flips, and critical point is where slope is 0

- dan815

and roots are where the function is 0

- Michele_Laino

I thin|dw:1435145496095:dw|k that x=0 is an inflection point with inclined tangent, like this:

- anonymous

And concavity is usually determined by the second derivative? So what does it mean that this second derivative is always 0?

- Michele_Laino

oops..
...I think that x=0 is an inflection point with inclined tangent, like this...

- 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

Oh, I totally misread the question! sorry

- anonymous

So how should I be testing for it being an inflection point?

- 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

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

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

|dw:1435146282570:dw|

- anonymous

I see! That makes sense

- 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

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

yes! that's right
you should get this:

- Michele_Laino

\[y - 5 = 2\left( {x - 5} \right)\]

- 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

oops...wrong equation. Here is the right equation
\[y - 5 = 2x\]

- 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

where does the g(0) come from?

- 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

where:
\[o\left( {x - {x_0}} \right)\]
is a quantity which goes to zero as x approaches to x_0

- anonymous

okay

- Michele_Laino

so substituting your data we can write:
\[g\left( {0.3} \right) = 5 + 0.6 = 5.6\]

- anonymous

Awesome! And then for g''(x)... does this look correct: 6x^2+6x+4
so that g''(0)=4

- Michele_Laino

Starting for m your formula for g '(x), after simple computation, I got this:

- 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

so, substituting x=0, we can write:
\[g''\left( 0 \right) = 4\]

- Michele_Laino

oops.. starting from your formula...

- anonymous

Thank you so much! This has been a great help!!

- 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

sorry for my error of computation

- anonymous

so g(.3) now is 8.6, yes?

- Michele_Laino

yes! we have:
\[g\left( {0.3} \right) = 5 + 12 \times 0.3 = 8.6\]

- anonymous

Thank you!!

- Michele_Laino

:)

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