Question

How do you use the first derivative test to determine the local extrema $f\left( x \right)=3{{x}^{5}}-20{{x}^{3}}$?

Answer 1

From the question given we have to find the local extrema of $f\left( x \right)=3{{x}^{5}}-20{{x}^{3}}$ by using first derivative test. As we know that the first derivative test allows us to find the critical points of a function by these points we have to check whether these points are local extrema by looking at the sign of the first derivative changes around them.

Complete step by step solution:
From the question we have to find the local extrema of the function using first derivative test
$\Rightarrow f\left( x \right)=3{{x}^{5}}-20{{x}^{3}}$
Now we have to do the first derivative of the given function then we will get,
$\Rightarrow {{f}^{'}}\left( x \right)=15{{x}^{4}}-60{{x}^{2}}$
To find the functions critical points, make the first derivative equal to zero and solve for x
By doing this we will get,
$\Rightarrow {{f}^{'}}\left( x \right)=15{{x}^{4}}-60{{x}^{2}}=0$
$\Rightarrow 15{{x}^{4}}-60{{x}^{2}}=0$
$\Rightarrow 15{{x}^{2}}\left( {{x}^{2}}-4 \right)=0$
$\Rightarrow 15{{x}^{2}}\left( x-2 \right)\left( x+2 \right)=0$
The equations have three solutions they are
$\Rightarrow x=0,x=\pm 2$
Since functions have no domain restrictions, all these points will be critical points.
Now, in order for the points to be local extrema, the function must go from increasing, i.e., a positive ${{f}^{'}}\left( x \right)$, to decreasing i.e., a negative ${{f}^{'}}\left( x \right)$ vice versa.
Since, we have three critical points, we have to look in the four intervals. We have to select a value from each interval to determine the sign of the first of the first derivative on that interval.
Case-1: $\left( -\infty ,-2 \right)$
By taking the value $-3$
$\Rightarrow {{f}^{'}}\left( -3 \right)=15{{\left( -3 \right)}^{4}}-60{{\left( -3 \right)}^{2}}$
$\Rightarrow {{f}^{'}}\left( -3 \right)=1215-540=675$
Therefore, it is a positive.
Case-2: $\left( -2,0 \right)$
By taking the value $-1$
$\Rightarrow {{f}^{'}}\left( -1 \right)=15{{\left( -1 \right)}^{4}}-60{{\left( -1 \right)}^{2}}$
$\Rightarrow {{f}^{'}}\left( -1 \right)=15-60=-45$
Therefore, it is a negative.
Case-3: $\left( 0,2 \right)$
By taking the value $1$
$\Rightarrow {{f}^{'}}\left( 1 \right)=15{{\left( 1 \right)}^{4}}-60{{\left( 1 \right)}^{2}}$
$\Rightarrow {{f}^{'}}\left( 1 \right)=15-60=-45$
Therefore, it is a negative.
Case-4: $\left( 2,\infty \right)$
By taking the value $3$
$\Rightarrow {{f}^{'}}\left( 3 \right)=15{{\left( 3 \right)}^{4}}-60{{\left( 3 \right)}^{2}}$
$\Rightarrow {{f}^{'}}\left( 1 \right)=1215-540=675$
Therefore, it is a positive.
We know that the first derivative change sign, i.e., it goes from positive to negative, around $-2$, so this point is a local maximum.
On the other hand, the first derivative does not change sign around point 0, so this point will not be of interest.
Further we will notice that the first derivative changes sign again around point $2$, since it goes from negative to positive. This means that point $2$will be a local minimum.
To get the actual points that match these critical points, evaluate the original function in $x=-2$ and $x=-2$ that is
$\Rightarrow f\left( -2 \right)=3{{\left( -2 \right)}^{5}}-20{{\left( -2 \right)}^{3}}=64$
$\Rightarrow f\left( 2 \right)=3{{\left( 2 \right)}^{5}}-20{{\left( 2 \right)}^{3}}=-64$

Therefore, the function $f\left( x \right)=3{{x}^{5}}-20{{x}^{3}}$ has a local minimum at $\left( 2,-64 \right)$ and a local maximum at $\left( -2,64 \right)$.

Note: Students should know the concept of the first derivative test, students should recall the concept of functions while doing the above problem. Students should be very careful while doing the differentiation and substitution.