How do I start this derivative problem?

The graphs of the function *f* (given in blue) and *g* (given in red) are plotted above. Suppose that *u*(*x*)=*f*(*x*)*g*(*x*) and *v*(*x*)=*f*(*x*)http://webwork.morris.umn.edu/webwor...144/char3D.png*g*(*x*). Find each of the following:

The graph: http://webwork.morris.umn.edu/webwor...rob6image1.png

Re: How do I start this derivative problem?

Find each of the following what? You already showed the graph, so you should be able to figure out $\displaystyle f(x)$ and $\displaystyle g(x)$ as piecewise continuous functions.

$\displaystyle f(x) = \left\{\begin{array}{ll}m_1 x + b_1 & -1<x<2 \\ m_2 x + b_2 & 2<x<6 \end{array} \right.$

$\displaystyle g(x) = \left\{\begin{array}{ll}m_3 x + b_3 & -1 < x < 0 \\ m_4 x + b_4 & 0 < x < 3 \\ m_5 x + b_5 & 3 < x < 6 \end{array} \right. $

Find the five slopes and five corresponding y-intercepts (to find the y-intercepts, you pretend that each line segment continues to figure out where it would hit the y-axis if it were a line). To be the most accurate, try to find specific points that the line segments cross. For example, the left-most red line segment passes between the points $\displaystyle (-1,5)$ and $\displaystyle (1,2)$, so you can use those points to find the slope of that line segment, which would tell you $\displaystyle m_1$.

What else do you need help figuring out?

Re: How do I start this derivative problem?

So how would I find u'(x) from u(x)? and v'(x) from v(x)?

Re: How do I start this derivative problem?

Quote:

Originally Posted by

**rhcprule3** So how would I find u'(x) from u(x)? and v'(x) from v(x)?

There are at least two ways of finding those derivatives. One way: use the product rule to find $\displaystyle u'(x)$ and the quotient rule to find $\displaystyle v'(x)$. Another way: write out the piecewise functions and differentiate. For example:

$\displaystyle u(x) = \left\{ \begin{array}{ll} (m_1 x + b_1)(m_3 x + b_3) & -1 < x < 0 \\ (m_1 x + b_1)(m_4 x + b_4) & 0 < x < 2 \\ (m_2 x + b_2)(m_4 x + b_4) & 2 < x < 3 \\ (m_2 x + b_2)(m_5 x + b_5) & 3 < x < 6\end{array} \right.$

And similarly for $\displaystyle v(x)$.

Re: How do I start this derivative problem?

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

u(1) =

v(1) =

this is the actual question

so to find u'(1), I would need to use the product rule?

Re: How do I start this derivative problem?

Quote:

Originally Posted by

**rhcprule3** The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

u(1) =

v(1) =

this is the actual question

so to find u'(1), I would need to use the product rule?

Now I am confused. Do you want u(1) or u'(1)?

Assuming you want u'(1), do you know the product rule? If you do, then find u'(x) and plug in 1.

Re: How do I start this derivative problem?

Quote:

Originally Posted by

**rhcprule3** The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

u(1) =

v(1) =

this is the actual question

so to find u'(1), I would need to use the product rule?

v(x)=f(x)g(x)

u(x)=f(x)g(x)

What is the distinction between u and v? Are you sure this is product problem or it is a composition problem, that is

u(x) = f(x) o g(x)

v(x) = g(x) o f(x)

Re: How do I start this derivative problem?

If I read it correctly (and there are some strange characters in the post), u= f **times** g and v= f **divided by** g.

Yes, by the product rule, u'= f'g+ fg' and by the quotient rule, v'= (f'g- fg')/v^2. Now, from the graph, f' is -3 for x< 0, 2 for 0< x< 3, and -1 for x> 3 while g' is -3/2 for x< 2 and 1/2 for x> 2.

Re: How do I start this derivative problem?

Quote:

Originally Posted by

**HallsofIvy** If I read it correctly (and there are some strange characters in the post), u= f **times** g and v= f **divided by** g.

Yes, by the product rule, u'= f'g+ fg' and by the quotient rule, v'= (f'g- fg')/v^2. Now, from the graph, f' is -3 for x< 0, 2 for 0< x< 3, and -1 for x> 3 while g' is -3/2 for x< 2 and 1/2 for x> 2.

I went backward to the first post, you got me, it looks like a division OP rewrite it as product in the fifth post, probably a typo. That confused me.