help with derivatives...

**mer1988** · November 16th 2007, 12:11 PM

The graph of a derivative f '(x) is shown in Figure 5.61.

ok, im given this graph of the derivative of f(x)...
can some help me find the coordinates to f(x)...i was given that f(0)=5
thanks...

mer1988

**Soroban** · November 16th 2007, 01:06 PM

Hello, mer1988!

This takes quite a bit of work (but simple stuff).
I will also assume that $f(x)$ is continuous.

The graph of a derivative $f'(x)$ is given.

i was given that: $f(0)\,=\,5$

We'll examine the function over the different intervals.

On $[0,\,1]$ , we see that: . $f'(x)\:=\:-x$
Integrate: . $f(x) \:=\:-\frac{1}{2}x^2 + C$
Since $(0,\,5)$ is on the graph: . $5 \:=\:-\frac{1}{2}\!\cdot\!0^2 + C\quad\Rightarrow\quad C\,=\,5$
Hence: on ${\color{blue}[0,\,1],\;f(x) \:=\:-\frac{1}{2}x^2 + 5}$
. . Note that: . $f(1) \:=\:\frac{9}{2}$

On $[1,3],\;f'(x) \:=\:-1$

Integrate: . $f(x)\:=\:-x + C$

Since $\left(1,\,\frac{9}{2}\right)$ is on the graph: . $\frac{9}{2}\:=\:-1 + C\quad\Rightarrow\quad C \:=\:\frac{11}{2}$
Hence: [color=blue}on [/color] ${\color{blue}[1,\,3],\;f(x) \:=\: -x + \frac{11}{2}}$
. . Note that: . $f(3) \,=\,\frac{5}{2}$

On $[3,\,5],\;f'(x) \:=\:x - 4$

Integrate: . $f(x)\;=\;\frac{1}{2}x^2 - 4x + C$
Since $\left(3,\,\frac{5}{2}\right)$ is on the curve: . $\frac{5}{2}\:=\:\frac{1}{2}\!\cdot\!3^1 - 4(3) + C\quad\Rightarrow\quad C \:=\:10$
Hence: on ${\color{blue}[3,\,5],\;f(x) \:=\:\frac{1}{2}x^2 - 4x + 10}$
. . Note that: . $f(5) = \frac{5}{2}$

On $[5,\,6],\;f'(x) \:=\:1$

Integrate: . $f(x) \:=\:x + C$

Since $\left(5,\,\frac{5}{2}\right)$ in on the curve: . $\frac{5}{2}\:=\:5 + C\quad\Rightarrow\quad C = -\frac{5}{2}$
Hence: on ${\color{blue}[5,\,6],\;f(x) \:=\:x - \frac{5}{2}}$

We have the piecewise function: . $f(x) ;=\;\left\{\begin{array}{ccc}-\frac{1}{2}x^2+5 & \;\; & 0 \leq x \leq 1 \\-x + \frac{11}{2} & \;\; & 1 \leq x \leq 3 \\\frac{1}{2}x^2-4x+10 & \;\; & 3 \leq x \leq 5 \\x - \frac{5}{2} & \;\; &5 \leq x \leq 6\end{array}\right\}$

And you can graph $f(x)$ now . . .

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