Help with Extreme Values!! Max. & Min.

[jessi88]

1) Find a Cubic function f(x) = ax^3 + bx^2 + cx + d
that has a local maximum at (-3, 3) and a local minimum
at (2,0).


2) Show that a cubic graph has 1 point of inflection (point of inflection - point where the curve changes concavity). If the graph has x- intercepts p, q, and r, show that the x - coordinates of the point of inflection is p + q + r / 3

Thanks

 

[pka]

For #2, you know the x-intercepts. Write the function in factored form.
\(\displaystyle \L
\begin{array}{l}
f(x) = (x - p)(x - q)(x - r) \\
f'(x) = (x - q)(x - r) + (x - p)(x - r) + (x - p)(x - r) \\
f''(x) = (x - r) + (x - q) + (x - r) + (x - p) + (x - r) + (x - p) \\
\end{array}\)
After finding the derivatives, we need to solve f’’(x)=0 to know the point of inflection.

 

[galactus]

1) Find a Cubic function f(x) = ax^3 + bx^2 + cx + d
that has a local maximum at (-3, 3) and a local minimum
at (2,0).

I reckon here's one way to attempt it.


\(\displaystyle \L\\a(-3)^{3}+b(-3)^{2}+c(-3)+d=3\rightarrow-27a+9b-3c+d=3\)

\(\displaystyle \L\\a(2)^{3}+b(2)^{2}+c(2)+d=0\rightarrow8a+4b+2c+d=0\)

Derivative of general function:

\(\displaystyle \L\\3ax^{2}+2bx+c\)

\(\displaystyle \L\\3a(-3)^{2}+2b(-3)+c=0\rightarrow27a-6b+c=0\)

\(\displaystyle \L\\3a(2)^{2}+2b(2)+c=0\rightarrow12a+4b+c=0\)

\(\displaystyle \L\\27a-6b+c=0\\
-27a+9b-3c+d=3\\
12a+4b+c=0\\
8a+4b+2c+d=0\)

\(\displaystyle \L\\\left[\begin{array}{cccc|c}27&-6&1&0&0\\{-27}&9&-3&1&3\\12&4&1&0&0\\8&4&2&1&0\end{array}\right]\)

Solve the system.

Try using the 2nd derivative test to see if they are, indeed, max and min at the designated values. If f''(-3)<0, then it is a local maximum; if f''(2)>0, then it is a local minimum