What are the points of inflection of $f (x) = 8 x^{2} + sin (2 x - π)$ $f(x)=8x^2 + sin(2x-pi)$ on $x \in [0, 2 π]$ $x in [0, 2pi]$ ?

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What are the points of inflection of $f (x) = 8 x^{2} + sin (2 x - π)$ $f(x)=8x^2 + sin(2x-pi)$ on $x \in [0, 2 π]$ $x in [0, 2pi]$ ?

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Answer 1 · 2018-04-30T03:47:31

None. No points of inflection.

Explanation:

Points of inflection are slope extrema or $f'(x)$ extrema. To find a maximum or minimum, take the second derivative and set equal to zero, to find where change in slope is zero.

$f(x)=8x^2+sin(2x-pi)$
$f'(x)=16x + 2cos(2x-pi)$
$f''(x)=16 - 4sin(2x-pi)$

Now set $f''(x)=0$ :

$0=16-4sin(2x-pi)$
$-16=-4sin(2x-pi)$
$4=sin(2x-pi)$
$sin^-1(4)=2x-pi$
$->$ NO SOLUTION

There are no (real, I dunno about complex) solutions as sin(x) cannot be greater than 1.

Therefore there are no points of inflection.

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What are the points of inflection of $f (x) = 8 x^{2} + sin (2 x - π)$ $f(x)=8x^2 + sin(2x-pi)$ on $x \in [0, 2 π]$ $x in [0, 2pi]$ ?

1 Answer

Explanation:

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What are the points of inflection of f(x)=8x^2 + sin(2x-pi) f(x)=8x2+sin(2x−π)f(x)=8x^2 + sin(2x-pi) on x in [0, 2pi]x∈[0,2π] x in [0, 2pi]?

1 Answer

Explanation:

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What are the points of inflection of $f (x) = 8 x^{2} + sin (2 x - π)$ $f(x)=8x^2 + sin(2x-pi)$ on $x \in [0, 2 π]$ $x in [0, 2pi]$ ?