inflection f(x)= 7/(x-7) — step-by-step solution

⇥1) Let's evaluate the domain of

\frac{7}{x - 7}

⇥ For the expression

\frac{7}{x - 7}

to be defined, we must have:

x - 7 \neq 0

⇥Solution obtained:

x \neq 7

An inflection point must lie on the curve: stay within the domain

x \in ℝ \ {7}

⇥To simplify differentiation, we rewrite

\frac{7}{x - 7}

7 \cdot {(x - 7)}^{-1}

⇥2) Let's differentiate:

\frac{d}{d x} (7 \cdot {(x - 7)}^{-1})

⇥We differentiate using:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{(n - 1)}

(where f =

x - 7

)

⇥

7 \cdot (-1 \cdot \frac{d}{d x} (x - 7) \cdot {(x - 7)}^{-2})

⇥ℹ

7 \cdot (-1)

-7

‖

\frac{d}{d x} (x - 7)

1 + 0

⇥

-7 \cdot (1 + 0) \cdot {(x - 7)}^{-2}

⇥

-7 \cdot {(x - 7)}^{-2}

⇥

- \frac{7}{{(x - 7)}^{2}}

⇥To simplify differentiation, we rewrite

- \frac{7}{{(x - 7)}^{2}}

-7 \cdot {(x - 7)}^{-2}

⇥3) Let's perform the differentiation:

\frac{d}{d x} (-7 \cdot {(x - 7)}^{-2})

⇥We differentiate using:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{(n - 1)}

(where f =

x - 7

)

⇥

-7 \cdot (-2 \cdot \frac{d}{d x} (x - 7) \cdot {(x - 7)}^{-3})

⇥ℹ

-7 \cdot (-2)

14

‖

\frac{d}{d x} (x - 7)

1 + 0

⇥

14 \cdot (1 + 0) \cdot {(x - 7)}^{-3}

⇥

14 \cdot {(x - 7)}^{-3}

⇥

\frac{14}{{(x - 7)}^{3}}

The second derivative

\frac{14}{{(x - 7)}^{3}}

is identically zero or never zero on each domain interval. The excluded pole is not an inflection point

▶Inflection points:

\emptyset