Jensen Mclean

2022-10-02

Why do I get different results when testing for increasing/decreasing intervals of a function here?

I have the function $f(x)=6+\frac{6}{x}+\frac{6}{{x}^{2}}$ and I want to find the intervals where it increases or decreases. The problem is that when I find ${f}^{\prime}(x)=0$, which becomes $x=-2$. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when $x>-2$. For example, ${f}^{\prime}(-1)>0$ and ${f}^{\prime}(5)<0$.

I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.

How I found the first derivative:

$\frac{d}{dx}\frac{6}{x}=\frac{[0]-[6\cdot 1]}{{x}^{2}}=\frac{-6}{{x}^{2}}$

$\frac{d}{dx}\frac{6}{{x}^{2}}=\frac{[0]-[6\cdot 2x]}{({x}^{2}{)}^{2}}=\frac{-12}{{x}^{3}}$

${f}^{\prime}(x)=\frac{-6}{{x}^{2}}-\frac{-12}{{x}^{3}}$

How I found the asymptotes:

When you combine the fractions of f(x), $f(x)=\frac{6{x}^{2}+6x+6}{{x}^{2}}$. When set equal to zero, ${x}^{2}$ has an x-value of zero. Therefore, f(x) has a vertical asymptote at $x=0$.

Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So f(x) has a horizontal asymptote at $y=6$.

How I found the value of ${f}^{\prime}(x)=0$:

$\frac{-6}{{x}^{2}}-\frac{12}{{x}^{3}}=0$

Add $\frac{12}{{x}^{3}}$ to both sides

$\frac{6}{{x}^{2}}=\frac{-12}{{x}^{3}}$

Multiply both sides by ${x}^{3}$.

$6x=-12$

Divide both sides by 6.

$x=\frac{-12}{6}=-2$

I have the function $f(x)=6+\frac{6}{x}+\frac{6}{{x}^{2}}$ and I want to find the intervals where it increases or decreases. The problem is that when I find ${f}^{\prime}(x)=0$, which becomes $x=-2$. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when $x>-2$. For example, ${f}^{\prime}(-1)>0$ and ${f}^{\prime}(5)<0$.

I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.

How I found the first derivative:

$\frac{d}{dx}\frac{6}{x}=\frac{[0]-[6\cdot 1]}{{x}^{2}}=\frac{-6}{{x}^{2}}$

$\frac{d}{dx}\frac{6}{{x}^{2}}=\frac{[0]-[6\cdot 2x]}{({x}^{2}{)}^{2}}=\frac{-12}{{x}^{3}}$

${f}^{\prime}(x)=\frac{-6}{{x}^{2}}-\frac{-12}{{x}^{3}}$

How I found the asymptotes:

When you combine the fractions of f(x), $f(x)=\frac{6{x}^{2}+6x+6}{{x}^{2}}$. When set equal to zero, ${x}^{2}$ has an x-value of zero. Therefore, f(x) has a vertical asymptote at $x=0$.

Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So f(x) has a horizontal asymptote at $y=6$.

How I found the value of ${f}^{\prime}(x)=0$:

$\frac{-6}{{x}^{2}}-\frac{12}{{x}^{3}}=0$

Add $\frac{12}{{x}^{3}}$ to both sides

$\frac{6}{{x}^{2}}=\frac{-12}{{x}^{3}}$

Multiply both sides by ${x}^{3}$.

$6x=-12$

Divide both sides by 6.

$x=\frac{-12}{6}=-2$

Marcel Mccullough

Beginner2022-10-03Added 11 answers

Step 1

Notice that ${f}^{\mathrm{\prime}}(x)=-\frac{6}{{x}^{2}}-\frac{12}{{x}^{3}}=-\frac{6}{{x}^{3}}(x+2)$.

So f′ is undefined at $x=0$ and f has slope 0 at $x=-2$.

Step 2

You should therefore check for increasing/decreasing on the intervals $(-\mathrm{\infty},-2),\phantom{\rule{thinmathspace}{0ex}}(-2,0)$ and $(0,\mathrm{\infty})$.

Notice that ${f}^{\mathrm{\prime}}(x)=-\frac{6}{{x}^{2}}-\frac{12}{{x}^{3}}=-\frac{6}{{x}^{3}}(x+2)$.

So f′ is undefined at $x=0$ and f has slope 0 at $x=-2$.

Step 2

You should therefore check for increasing/decreasing on the intervals $(-\mathrm{\infty},-2),\phantom{\rule{thinmathspace}{0ex}}(-2,0)$ and $(0,\mathrm{\infty})$.

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