Let f(x) = \frac{x^5}{3 + 1}. The inflection point (i.e. (x,y) coordinates) is: choose from the...


Let {eq}f(x) = \frac{x^5}{3 + 1} {/eq}. The inflection point (i.e. (x,y) coordinates) is: choose from the following:

a) (-1,0)

b) (0,1)

c) y = 1

d) x = 0

e) (0,-1)

f) (0,0)

g) y = 0

h) There is no inflection point.

i) (1,0)

j) x = 1

Inflection Points:

Recall that an inflection point is a point on the graph where the concavity changes. A curve is concave up when the second derivative is equal to 0, and concave down when the second derivative is negative. Inflection points occur where the second derivative equals 0.

Answer and Explanation:

This function is very simple and written strangely; most likely there is an error here, but that just makes things easier for us! We have

{eq}\begin{align*} f (x) &= \frac{x^5}{3+1 } = \frac14 x^5 \end{align*} {/eq}


{eq}\begin{align*} f' (x) &= \frac{d}{dx} \left( \frac14 x^5 \right) \\ &= \frac54 x^4 \end{align*} {/eq}

and then

{eq}\begin{align*} f '' (x) &= \frac{d}{dx} \left( \frac54 x^4 \right) \\ &= 5 x^3 \end{align*} {/eq}

This is clearly equal to 0 when {eq}x = 0 {/eq} (it is in fact a triple root). So the inflection point is the origin: (0,0), which is option f.

Note that the answer cannot be (d) because {eq}x = 0 {/eq} is a line, not a point.

Learn more about this topic:

Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5

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