Question

8). use the first derivative test to find the local maxima and minima for f(x) = 6x⁴ln(x²) − 2x⁴. show all your work.

Explanation:

Step1: Simplify the function

First, use the logarithm property \(\ln(x^{2}) = 2\ln|x|\) (for \(x eq0\)). So the function \(f(x)=6x^{4}\cdot2\ln|x|-2x^{4}=12x^{4}\ln|x|-2x^{4}\). The domain of \(f(x)\) is \(x\in(-\infty,0)\cup(0,+\infty)\) since \(\ln|x|\) is undefined at \(x = 0\).

Step2: Find the derivative \(f'(x)\)

Use the product rule \((uv)'=u'v + uv'\) and the power rule \((x^{n})'=nx^{n - 1}\), and the derivative of \(\ln|x|\) is \(\frac{1}{x}\) (for \(x eq0\)).
Let \(u = 12x^{4}\) and \(v=\ln|x|\), and \(w=-2x^{4}\).
First, find the derivative of \(12x^{4}\ln|x|\):
\(u' = 48x^{3}\), \(v'=\frac{1}{x}\) (for \(x eq0\)). So \((12x^{4}\ln|x|)'=48x^{3}\ln|x|+12x^{4}\cdot\frac{1}{x}=48x^{3}\ln|x| + 12x^{3}\).
The derivative of \(-2x^{4}\) is \(-8x^{3}\).
So \(f'(x)=48x^{3}\ln|x|+12x^{3}-8x^{3}=48x^{3}\ln|x| + 4x^{3}\).
Factor out \(4x^{3}\): \(f'(x)=4x^{3}(12\ln|x| + 1)\).

Step3: Find critical points

Set \(f'(x) = 0\). So \(4x^{3}(12\ln|x|+1)=0\).
This gives two cases:
- Case 1: \(4x^{3}=0\) \(\Rightarrow\) \(x = 0\), but \(x = 0\) is not in the domain of \(f(x)\), so we discard it.
- Case 2: \(12\ln|x|+1 = 0\) \(\Rightarrow\) \(\ln|x|=-\frac{1}{12}\) \(\Rightarrow\) \(|x|=e^{-\frac{1}{12}}\) \(\Rightarrow\) \(x=\pm e^{-\frac{1}{12}}\).

Step4: Apply the First Derivative Test

We divide the domain into intervals based on the critical points \(x=-e^{-\frac{1}{12}}\) and \(x = e^{-\frac{1}{12}}\). The intervals are \((-\infty,-e^{-\frac{1}{12}})\), \((-e^{-\frac{1}{12}},0)\), \((0,e^{-\frac{1}{12}})\), and \((e^{-\frac{1}{12}},+\infty)\).

For \(x\in(-\infty,-e^{-\frac{1}{12}})\):

Let's pick a test point, say \(x=-1\) (since \(-1<-e^{-\frac{1}{12}}\) as \(e^{-\frac{1}{12}}\approx0.92\)).
\(x^{3}=-1<0\), \(\ln|x|=\ln(1) = 0\), so \(12\ln|x|+1=1>0\). Then \(f'(x)=4x^{3}(12\ln|x| + 1)=4\times(-1)\times1=-4<0\). So \(f(x)\) is decreasing on \((-\infty,-e^{-\frac{1}{12}})\).

For \(x\in(-e^{-\frac{1}{12}},0)\):

Pick a test point, say \(x=-e^{-\frac{1}{24}}\) (since \(-e^{-\frac{1}{12}}<-e^{-\frac{1}{24}}<0\)).
\(x^{3}<0\) (because \(x\) is negative), \(\ln|x|=\ln(e^{-\frac{1}{24}})=-\frac{1}{24}\), so \(12\ln|x|+1=12\times(-\frac{1}{24})+1=-0.5 + 1 = 0.5>0\). Then \(f'(x)=4x^{3}(12\ln|x| + 1)\), since \(x^{3}<0\) and \(12\ln|x| + 1>0\), \(f'(x)<0\)? Wait, no: \(x\in(-e^{-\frac{1}{12}},0)\), \(x\) is negative, so \(x^{3}\) is negative. \(12\ln|x|+1\): when \(x\in(-e^{-\frac{1}{12}},0)\), \(|x|\in(0,e^{-\frac{1}{12}})\), so \(\ln|x|\in(-\frac{1}{12},-\infty)\), \(12\ln|x|+1\in(0,12\times(-\infty)+1)\)? Wait, no, when \(x=-e^{-\frac{1}{12}}\), \(\ln|x|=-\frac{1}{12}\), \(12\ln|x|+1 = 0\). For \(x\in(-e^{-\frac{1}{12}},0)\), \(|x|<e^{-\frac{1}{12}}\), so \(\ln|x|<-\frac{1}{12}\), so \(12\ln|x|+1<0\). Ah, I made a mistake earlier. Let's recalculate:
If \(x\in(-e^{-\frac{1}{12}},0)\), then \(|x|\in(0,e^{-\frac{1}{12}})\), so \(\ln|x|<\ln(e^{-\frac{1}{12}})=-\frac{1}{12}\), so \(12\ln|x|+1<12\times(-\frac{1}{12})+1=-1 + 1 = 0\). So \(12\ln|x|+1<0\) in this interval. And \(x^{3}<0\) (since \(x\) is negative), so \(4x^{3}(12\ln|x| + 1)\): negative times negative is positive. So \(f'(x)>0\) on \((-e^{-\frac{1}{12}},0)\).

For \(x\in(0,e^{-\frac{1}{12}})\):

Pick a test point, say \(x = e^{-\frac{1}{24}}\) (since \(0<e^{-\frac{1}{24}}<e^{-\frac{1}{12}}\)).
\(x^{3}>0\) (since \(x\) is positive), \(\ln|x|=\ln(e^{-\frac{1}{24}})=-\frac{1}{24}\), so \(12\ln|x|+1=12\times(-\frac{1}{24})+1=-0.5 + 1 = 0.5>0\)? Wait, no: \(x\in(0,e^{-\frac{1}{12}})\), \(|x|\in(0,e^{-\frac{1}{12}})\), so \(\ln|x|<-\frac{1}{12}\), so \(12\ln|x|+1<0\). So \(12\ln|x|+1<0\) in this interval. \(x^{3}>0\), so \(4x^{3}(12\ln|x| + 1)<0\). So \(f'(x)<0\) on \((0,e^{-\frac{1}{12}})\).

For \(x\in(e^{-\frac{1}{12}},+\infty)\):

Pick a test point, say \(x = 1\) (since \(1>e^{-\frac{1}{12}}\approx0.92\)).
\(x^{3}>0\), \(\ln|x|=\ln(1)=0\), so \(12\ln|x|+1 = 1>0\). So \(f'(x)=4x^{3}(12\ln|x| + 1)>0\). So \(f(x)\) is increasing on \((e^{-\frac{1}{12}},+\infty)\).

Now, analyze the sign changes of \(f'(x)\) at the critical points:

- At \(x=-e^{-\frac{1}{12}}\):
- For \(x < -e^{-\frac{1}{12}}\) (e.g., \(x=-1\)), \(f'(x)<0\) (function decreasing).
- For \(x\in(-e^{-\frac{1}{12}},0)\), \(f'(x)>0\) (function increasing).
So by the First Derivative Test, \(f(x)\) has a local minimum at \(x=-e^{-\frac{1}{12}}\) because the function changes from decreasing to increasing.

- At \(x = e^{-\frac{1}{12}}\):
- For \(x\in(0,e^{-\frac{1}{12}})\), \(f'(x)<0\) (function decreasing).
- For \(x>e^{-\frac{1}{12}}\) (e.g., \(x = 1\)), \(f'(x)>0\) (function increasing).
So by the First Derivative Test, \(f(x)\) has a local minimum at \(x = e^{-\frac{1}{12}}\) because the function changes from decreasing to increasing. Wait, wait, no: wait, when \(x\) approaches \(e^{-\frac{1}{12}}\) from the left (in \((0,e^{-\frac{1}{12}})\)), \(f'(x)<0\) (decreasing), and from the right (in \((e^{-\frac{1}{12}},+\infty)\)), \(f'(x)>0\) (increasing). So it's a local minimum. But wait, let's check the function values or re - check the derivative sign.

Wait, let's re - evaluate the sign of \(12\ln|x|+1\) in different intervals:

The equation \(12\ln|x|+1 = 0\) gives \(\ln|x|=-\frac{1}{12}\), so \(|x|=e^{-\frac{1}{12}}\).

- For \(|x|>e^{-\frac{1}{12}}\) (i.e., \(x\in(-\infty,-e^{-\frac{1}{12}})\cup(e^{-\frac{1}{12}},+\infty)\)):
- If \(x\in(-\infty,-e^{-\frac{1}{12}})\), \(|x|>e^{-\frac{1}{12}}\), so \(\ln|x|>-\frac{1}{12}\), then \(12\ln|x|+1>0\). And \(x^{3}<0\) (since \(x<0\)), so \(f'(x)=4x^{3}(12\ln|x| + 1)<0\).
- If \(x\in(e^{-\frac{1}{12}},+\infty)\), \(|x|>e^{-\frac{1}{12}}\), so \(\ln|x|>-\frac{1}{12}\), then \(12\ln|x|+1>0\). And \(x^{3}>0\) (since \(x>0\)), so \(f'(x)=4x^{3}(12\ln|x| + 1)>0\).

- For \(|x|<e^{-\frac{1}{12}}\) (i.e., \(x\in(-e^{-\frac{1}{12}},0)\cup(0,e^{-\frac{1}{12}})\)):
- If \(x\in(-e^{-\frac{1}{12}},0)\), \(|x|<e^{-\frac{1}{12}}\), so \(\ln|x|<-\frac{1}{12}\), then \(12\ln|x|+1<0\). And \(x^{3}<0\) (since \(x<0\)), so \(f'(x)=4x^{3}(12\ln|x| + 1)>0\) (negative times negative is positive).
- If \(x\in(0,e^{-\frac{1}{12}})\), \(|x|<e^{-\frac{1}{12}}\), so \(\ln|x|<-\frac{1}{12}\), then \(12\ln|x|+1<0\). And \(x^{3}>0\) (since \(x>0\)), so \(f'(x)=4x^{3}(12\ln|x| + 1)<0\) (positive times negative is negative).

Now, the sign changes:

- At \(x=-e^{-\frac{1}{12}}\):
- Left of \(x=-e^{-\frac{1}{12}}\) (e.g., \(x=-2\)): \(f'(x)<0\) (decreasing).
- Right of \(x=-e^{-\frac{1}{12}}\) (e.g., \(x=-e^{-\frac{1}{24}}\)): \(f'(x)>0\) (increasing).
So by the First Derivative Test, \(x=-e^{-\frac{1}{12}}\) is a local minimum.

- At \(x = e^{-\frac{1}{12}}\):
- Left of \(x = e^{-\frac{1}{12}}\) (e.g., \(x=e^{-\frac{1}{24}}\)): \(f'(x)<0\) (decreasing).
- Right of \(x = e^{-\frac{1}{12}}\) (e.g., \(x = 2\)): \(f'(x)>0\) (increasing).
So by the First Derivative Test, \(x = e^{-\frac{1}{12}}\) is a local minimum.

Now, let's find the local minimum values.

First, simplify \(f(x)\) for \(x>0\) (since \(f(x)\) is even? Let's check: \(f(-x)=6(-x)^{4}\ln((-x)^{2})-2(-x)^{4}=6x^{4}\ln(x^{2})-2x^{4}=f(x)\), so \(f(x)\) is an even function. So \(f(-e^{-\frac{1}{12}})=f(e^{-\frac{1}{12}})\).

Let's compute \(f(e^{-\frac{1}{12}})\):

\(f(x)=12x^{4}\ln|x|-2x^{4}\), for \(x = e^{-\frac{1}{12}}\), \(\ln|x|=-\frac{1}{12}\).

\(f(e^{-\frac{1}{12}})=12(e^{-\frac{1}{12}})^{4}\times(-\frac{1}{12})-2(e^{-\frac{1}{12}})^{4}\)

\(=(e^{-\frac{1}{3}})\times(-1)-2e^{-\frac{1}{3}}=-3e^{-\frac{1}{3}}\)

So the local minima occur at \(x = \pm e^{-\frac{1}{12}}\) and the local minimum value is \(-3e^{-\frac{1}{3}}\). There are no local maxima because the function changes from decreasing to increasing at both critical points (so both are minima) and the function tends to \(+\infty\) as \(x ightarrow\pm\infty\) (since the leading term of \(f(x)\) is \(6x^{4}\ln(x^{2})\) which behaves like \(12x^{4}\ln|x|\), and as \(x ightarrow\pm\infty\), \(x^{4}\) grows faster than \(\ln|x|\) decays, so \(f(x) ightarrow+\infty\)) and there are no points where the function changes from increasing to decreasing.

Answer:

The function \(f(x) = 6x^{4}\ln(x^{2})-2x^{4}\) has local minima at \(x = \pm e^{-\frac{1}{12}}\) with local minimum value \(f(\pm e^{-\frac{1}{12}})=-3e^{-\frac{1}{3}}\). There are no local maxima.