Write a linear function with f with f (5) = -1 and f (0) = -5
1 answer:
You might be interested in
We start with 
and wish to write it as 
First, pull 2 out from the first two terms:
Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have
and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square: 
The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have
and when we multiply that out it does not give us what we started with. It gives us 
So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.
We do this as follows:
which gives us the final expression we seek:
If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e = -103
We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106
First, pull 2 out from the first two terms:
Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have
The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have
So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.
We do this as follows:
If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e = -103
We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106
(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where
then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,
(b) The velocity after 3 seconds is
(c) The particle is at rest when its velocity is zero:
(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:
In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.
(e) The total distance traveled is given by the definite integral,
By definition of absolute value, we have
In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as
and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to