How do you find the local extrema for #f(x) = 3x^5 - 10x^3 - 1# on the interval [-1,1]?

Question

Scarlett Allison · Accepted Answer

That function has no local extrema on that interval. (It does, of course, have global extrema on the closed interval.)

I take the definition of local maximum to be: #f(c)# is a local maximum for #f# if and only if the is an open interval, #I#, containing #c# such that #f(c) >= f(x)# for all #x in I# (For local minimum, reverse the inequality.) Fermat's Theorem tells us that #f(c)# is a local extremum if and only if #f'(c) = 0# or #f'(c)# does not exist. For #f(x) = 3x^5-10x^3-1#, we have #f'(x) = 15x^4-30x^2 = 15x^2(x^2-2)#. #f'# never fails to exist and it is #0# at #x = 0#, #-sqrt2# and #sqrt2#. But the only one of the in the interval is #x = 0#. Applying the first derivative test, we see that #f(0)# is neither a minimum nor a maximum. (Near #0# on either side, we have #f'(x) < 0#, so #f# is decreasing on both sides of #0#.) Alternatives If you are being graded based on your answer, check you grader's definitions of local extrema. Some presentations of calculus may include clauses in the definition so that we need only an open interval, #I#, such that for all #x# in the intersection of #I# and the domain of #f#, we have #f(c) >= x# In such a presentation, since we have restricted the domain of #f# to #[-1,1]#, we must answer that #f(1) = -8# in a local minimum and #f(-1) = 6# is a local maximum.

Charles Arocha · Answer

undefined To find the local extrema of $f(x) = 3x^5 - 10x^3 - 1$ on the interval $[-1,1]$, follow these steps:

1. Find the critical points by setting the derivative equal to zero and solving for $x$.
2. Evaluate $f(x)$ at these critical points and at the endpoints of the interval.
3. The maximum and minimum values within the interval will be the local extrema.

Now, let's proceed with these steps:

1. Find the derivative of $f(x)$:

$$ f'(x) = 15x^4 - 30x^2 $$

Set $f'(x) = 0$ and solve for $x$:

$$ 15x^4 - 30x^2 = 0 $$

$$ 15x^2(x^2 - 2) = 0 $$

$$ x = 0, \pm \sqrt{2} $$

2. Evaluate $f(x)$ at the critical points and endpoints:

For $x = 0$, $f(0) = -1$.

For $x = \pm \sqrt{2}$, calculate $f(\sqrt{2})$ and $f(-\sqrt{2})$.

For $x = 1$, $f(1) = -8$.

For $x = -1$, $f(-1) = -12$.

3. Compare these values to determine the local extrema.

Local maximum: $f(-1) = -12$  
Local minimum: $f(\sqrt{2})$  
Local maximum: $f(0) = -1$  
Local minimum: $f(1) = -8$

So, the local extrema for $f(x)$ on the interval $[-1,1]$ are:  
Local maximum at $x = -1$ with a value of -12.  
Local minimum at $x = \sqrt{2}$ with a value to be calculated.  
Local maximum at $x = 0$ with a value of -1.  
Local minimum at $x = 1$ with a value of -8.

Now, calculate $f(\sqrt{2})$:

$$ f(\sqrt{2}) = 3(\sqrt{2})^5 - 10(\sqrt{2})^3 - 1 $$

$$ f(\sqrt{2}) = 24\sqrt{2} - 20\sqrt{2} - 1 $$

$$ f(\sqrt{2}) = 4\sqrt{2} - 1 $$

Therefore, the local minimum at $x = \sqrt{2}$ has a value of $4\sqrt{2} - 1$.