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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions -- Discussion 6}
\centerline{November 8, 2000}
\bigskip
\noindent
E.  The problem said to find all local maxima and minima of the two
functions.
\bigskip
\item{a)}  $f(x) = x^2e^{-2x}$.  Differentiating by the product and
chain rules, we have the first derivative 
$$f'(x) = x^2e^{-2x}(-2) + 2xe^{-2x} = (2x - 2x^2)e^{-2x}\leqno(1)$$
Factoring the polynomial in front, we also can write $f'(x)$
as 
$$f'(x) = 2x(1 - x)e^{-2x}\leqno(2)$$
Continuing from (1), using the product and chain rules again,
the second derivative is:
$$f''(x) = (2x - 2x^2)e^{-2x}(-2) + (2 - 4x)e^{-2x} = (2-8x+4x^2)e^{-2x}$$
From the factored form (2), the first derivative is zero at $x = 0$ and 
$x = 1$. 

Moreover:
$$f'(x) \cases{ < 0 & if $x < 0$\cr
                > 0 & if $0 < x < 1$\cr
                < 0 & if $x > 1$\cr}$$
                
\item{}                
Since $f$ is decreasing to the left of $x = 0$ and increasing to the right
of $x=0$ (until $x = 1$), this says that $x = 1$ is a local minimum for 
$f$.  Similarly, since $f$ is increasing to the left of $x = 1$ and decreasing to the right
of $x=1$,  $x = 1$ must be a local maximum.  (This is the First Derivative
Test.)  

\item{}
Using $f''(x)$, we see that $f''(0) = 2 > 0$ and $f''(1) = -2e^{-2} < 0$.
This also shows that $x = 0$ is a local minimum and $x = 1$ is a 
local maximum (Second Derivative Test).

\bigskip
\item{b)}  $f(x) = x^4 - 6x^2 + 1$.  Differentiating
we have the first derivative 
$$f'(x) = 4x^3 - 12x = 4x(x^2 - 3)\leqno(3)$$
Continuing from (1), 
the second derivative is:
$$f''(x) = 12x^2 - 12$$
From (3), the first derivative is zero at $x = 0$ and 
$x = \pm\sqrt{3}$. 

Moreover:
$$f'(x) \cases{ < 0 & if $x < -\sqrt{3}$\cr
                > 0 & if $-\sqrt{3} < x < 0$\cr
                < 0 & if $0 < x < \sqrt{3}$\cr
                > 0 & if $x > \sqrt{3}$\cr}$$
                
\item{}
By the First Derivative
Test, $x = 0$ is a local maximum and $x = \pm\sqrt{3}$ are local 
minima. 

\item{}
Using $f''(x)$, we see that $f''(0) =  -12 < 0$ and $f''(\pm\sqrt{3}) = 24 > 0$.
This also shows that $x = 0$ is a local maximum and $x = \pm\sqrt{3}$ are
local minima (Second Derivative Test).
\bye