how to find inflection point of f(x)=4x^2 +2x +1 ?

Question

marissalovescats · Answer

You will have to find the 2nd derivative, set it = to 0 and solve for x

marissalovescats · Answer

Or graph it and see where it crosses the x axis at.

anonymous · Answer

the second derivative is 8 though.

anonymous · Answer

so im guessing there is no inflection point ? o-o

imstuck · Answer

It's a parabola so it does have an inflection point, but the graph does not go through the x axis, so the solutions are imaginary.  The graph has a min value at the inflection point.

anonymous · Answer

is there a relative minimum at -1/4 ?

imstuck · Answer

The min of the graph occurs at (0, 1).

imstuck · Answer

That is where the graph reaches its lowest point and then begins to go back up.

anonymous · Answer

no.

imstuck · Answer

No what?

anonymous · Answer

i just graphed it.

imstuck · Answer

So did I.

anonymous · Answer

oh Lol

anonymous · Answer

well use this it will help you http://www.wolframalpha.com/input/?i=4x%5E2+%2B2x+%2B1&dataset=

jim_thompson5910 · Answer

IMStuck, all parabolas are either concave up or concave down throughout their entire domain. Since you have no change in concavity, you won't have an inflection point on a parabola.

anonymous · Answer

@IMStuck the graph reaches its lowest point at -1/4, 3/4 not 0,1 you graphed it wrong ?

anonymous · Answer

@jim_thompson5910 thank you... so it has no inflection point right..?

jim_thompson5910 · Answer

correct

here's how you show it through derivatives

f(x) = 4x^2 + 2x + 1

f ' (x) = 8x + 2

f '' (x) = 8

The second derivative is always positive. So f(x) is always concave up.

There are no changes in concavity (ie f ''(x) doesn't change in sign at all), so there are no inflection points.

anonymous · Answer

@jim_thompson5910 thanks for clarifying, also there is a relative minimum at -1/4 ? o-o

jim_thompson5910 · Answer

f ' (x) = 8x + 2

0 = 8x + 2

-2 = 8x

x = -2/8

x = -1/4

This is a critical value.

Because f ''(x) > 0 for all x, this means f(x) is concave up which leads to a relative min (turns out it's actually an absolute min too)

jim_thompson5910 · Answer

So yes, you are correct

anonymous · Answer

Okay thanks, that other guy confused me I thought I got it wrong!

jim_thompson5910 · Answer

you're welcome