Bapilievolia0o0

2023-02-18

Where does the graph of $y=\left(5{x}^{4}\right)-\left({x}^{5}\right)$ have an inflection point?

Dakota George

Beginner2023-02-19Added 9 answers

An inflection point on a graph is a point where the concavity changes. We'll look at the sign of the second derivative to investigate concavity:

$y=5{x}^{4}-{x}^{5}$

$y\prime =20{x}^{3}-5{x}^{4}$

$y\prime \prime =60{x}^{2}-20{x}^{3}=20{x}^{2}(3-x)$

Obviously $20{x}^{2}$ is always positive, so the sign of y'' is the same as the sign of $3-x$.

Which is positive for $x<3$ and negative for $x>3$. At $x=3$ the concavity changes.

Because an inflection point is a point on a graph, we need:

when $x=3$, we get

$y=5\left({3}^{4}\right)-{3}^{5}=5\left({3}^{4}\right)-3\left({3}^{4}\right)=2\left({3}^{4}\right)=2\left(81\right)=162$

The point (3, 162) in the only inflection point for the graph of $y=5{x}^{4}-{x}^{5}$

$y=5{x}^{4}-{x}^{5}$

$y\prime =20{x}^{3}-5{x}^{4}$

$y\prime \prime =60{x}^{2}-20{x}^{3}=20{x}^{2}(3-x)$

Obviously $20{x}^{2}$ is always positive, so the sign of y'' is the same as the sign of $3-x$.

Which is positive for $x<3$ and negative for $x>3$. At $x=3$ the concavity changes.

Because an inflection point is a point on a graph, we need:

when $x=3$, we get

$y=5\left({3}^{4}\right)-{3}^{5}=5\left({3}^{4}\right)-3\left({3}^{4}\right)=2\left({3}^{4}\right)=2\left(81\right)=162$

The point (3, 162) in the only inflection point for the graph of $y=5{x}^{4}-{x}^{5}$

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