Thread: interpreting the graph of the derivative

1. interpreting the graph of the derivative

graph of f '(x) is attached.
is this correct:
f is increasing on (3,6) and (6,8)
f is concave down on (0,1.5) and (4.5,6) and (6,9)
the critical numbers of f are x=0, x=3, x=6, x=8
the values of x that f has a local maximum is x=8
the values of x that f has a local minimum is x=3

2. Hello, Harry!

3. f is increasing on (3,6) and (6,8)
Good.

f is concave down on (0,1.5) and (4.5,6) and (6,9)
Why only "9]" and not "10]"?
No need to mention "convcave up"?

the critical numbers of f are x=0, x=3, x=6, x=8
What's critical about x = 0 that is not critical about x = 10?

the values of x that f has a local maximum is x=8
I'm always torn on this issue. Does "Local" include "Global"?
You may wish to ponder x = 0 on this one.

the values of x that f has a local minimum is x=3
What do you think of x = 10?

4. does anyone know what the graph of f would look like?

5. What's critical about x = 0 that is not critical about x = 10?
aren't critical numbers where f '(x)=0?

I'm always torn on this issue. Does "Local" include "Global"?
You may wish to ponder x = 0 on this one.
i think just "local"

What do you think of x = 10?
can endpoints be considered local maxima or local minima?

6. Hello, Harry!

Does anyone know what the graph of $f$ would look like?
Do you really want to know? . . . You may be sorry you asked!
First of all, I am guessing that $f'(x)$ consists of parabolic arcs.

On the interval $[0,\,3],\;f'(x)$ is a parabola through $(0,\,0),\:\left(\frac{3}{2},\:-1\right),\:(3,\,0)$
. . Its equation is: . $f'(x) \:=\:\frac{4}{9}x^2-\frac{4}{3}x$
Hence: . ${\color{blue}f(x) \;=\;\frac{4}{27}x^3 - \frac{2}{3}x^2 + C_1}$

On the interval $[3,\,6],\;f'(x)$ is a parabola through $(3,\,0),\:\left(\frac{9}{2},\,1\right),\:(6,\,0)$
. . Its equation is: . $f'(x)\:=\:-\frac{4}{9}x^2 + 4x - 8$
Hence: . ${\color{blue}f(x) \;=\;-\frac{4}{27}x^2 + 2x^2 - 8x + C_2}$

On the interval $[6,\,\infty],\;f'(x)$ is a parabola through $(6,\,2),\:(8,\,0)$
. . Assuming (6,2) is the vertex, its equation is: . $f'(x) \:=\:-\frac{1}{2}x^2 + 6x - 16$
Hence: . ${\color{blue}f(x) \;=\;-\frac{1}{6}x^3 + 3x^2 - 16x + C_3}$

We have a piecewise function: . $f(x) \;=\;\begin{Bmatrix}\frac{4}{27}x^3-\frac{2}{3}x^2+C_1 & & 0 \leq x \leq 3 \\ \\
-\frac{4}{27}x^3 +2x^2-8x + C_2 & & 3 \leq x \leq 6\\ \\
-\frac{1}{6}x^3 + 3x^2 - 16x + C_3 & & x \geq 6

\end{Bmatrix}$

7. Originally Posted by harry1
aren't critical numbers where f '(x)=0?
That definition is no good. The derivative has to exist in order for that to be sufficient.

What say you of x = 0 for f(x) = |x|? No derivative, but definitely a critical value.

Endpoints are the same. There is no derivative and they must be considered separately. Your text book or teacher should provide VERY CLEAR guidance on how to handle these situations.