I got this question in my mailbox recently:
I am having a problem with show my son how to work out the proportion problem and solving equation by division, like 7.y=105 64=y.8 and find the missing term for the proportion of 3/4=x/16.
Can you help me out?
Always remember this problem solving strategy: when a problem is too difficult, make another, similar, but in some way easier problem, and observe for a strategy to solve that one.
If 7 ⋅ y = 105 and 64 = y ⋅ 8 are difficult, use easier examples first. Have him solve these ones:
2 ⋅ __ = 4
5 ⋅ x = 10
2 ⋅ x = 6
and
8 = 2 ⋅ __
12 = 3 ⋅ x
9 = 3 ⋅ x
If 'x' intimidates him, you can use an empty line.
Then ask him, HOW did he solve these ones? Well, chances are, of course, that he just 'sees' the answer, or remembers his multiplication tables and gets the answer fro! m those.
But then SHOW him how division, in each case, gives us the answer too:
5 ⋅ x = 10. (We already know the answer is 2)
10 ÷ 5 gives the answer.
After going thru this, the initial problems should not be difficult. If they still do, then your son might have problems in understanding division concept and might need review in that area first.
AP EXAM PRACTICE QUIZ 1
Let h b! e a function defined for all x does not equal 0 such that h(4) = -3 and the derivative of h is given by h'(x) = (x^2-2)/x for all x does not equal 0.
a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values.
To determine if there are local extrema at the critical numbers, we must first determine the critical numbers. Remember that there are critical numbers when a function has an asymptote or is undefined. h' = 0 when x = sqrt(2) and -sqrt(2). h' is undefined at x = 0 since 0 isn't in the domain (as stated in the question). So critical numbers are x = sqrt(2), -sqrt(2), and 0.
We use the first derivative test on h' to see where h' is positive or negative. Why? Because when h' is positive, h is increasing; when h' is negative, h' is decreasing. So a change in sign in h' would indicate a slope of ze! ro at that point and that's where there are local extrema. Acc! ording t o the line analysis, we see that to the left of -sqrt(2), h' is negative; between -sqrt(2) and 0, h' is positive; between 0 and sqrt(2), h' is negative; and to the right of sqrt(2), h' is positive. Wherever h' changes sign from negative to positive, h has a local minimum; wherever h' changes sign from positive to negative, h has a local maximum. By the first derivative test (line analysis), there are local minimums at x = -sqrt(2) and x = sqrt(2). We don't look at 0 because it's not part of the domain of the function.
Rememberize these rules (from chapter 5 of your textbook):
If the second derivative is positive, the first derivative is increasing, and the parent function is concave up.
If the second derivative is negative, the first derivative is decreasing, and the parent function is concave down.
So wherever h" is positive, h is concave up.
We determine h" by differentiating h using the quotient rule.
We see that h" is positive, so h is concave up everywhere.
c) Write an equation for the line tangent to the graph of h at x = 4.
Pull out the point-slope formula: y-y1=m(x-x1)
The question gave us the x-coordinate: x = 4.
The question! gave us the y-coordinate: y= -3.
m is the slope at x = 4! , so plu g x = 4 into h' which spits out 7/2.
Plug those numbers into the equation. BING! BANG! BOOM! We're done part c.
y+3=(7/2)(x-4)
d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x > 4?
If we draw a line tangent to the graph at x = 4, the line is below the graph at x > 4, because h is concave up everywhere.
INTERMEDIATE VALUE THEOREM OF INTEGRALS
Similar to the intermediate ! (or mean) value theorem of derivatives (that in a closed interval between a and b there exists a point on a continuous function which equals the average value), there exists a point on a continuous function which equals the average integral of the function.
We have b = 3. We have a = 0. We have f(x) = 1-2x. Plug the numbers into the equation. BING! BANG! BOOM! We get the average integral. But why does it work?
< img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 400px; height: 361px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMJyur_sqimRoWoyFhF7L763IQ9-1sauBZeeEKWnFA7YGcITIgwIZcFLv1scAnXvQSR6hs3irI6Z5NyZceEvVAVFYpqT9FU0VoNb3tkG5TiIEB9LKZxKQR-XcxhL3yIvWi_8AHevWFspdM/s400/6.bmp" alt="" id="BLOGGER_PHOTO_ID_5310238211959744338" border="0" />In this graph, pivot the yellow area found between c and b to the white area found between c and a. Notice that they fit together like a jigsaw puzzle to yield one big yellow rectangle. We can imagine that one big rectangle having the same area as the area under the graph, which we can see as the average value of the function.
HOUSEKEEPING
I am hoping you can help me. I can not remember how to solve mixture problems and how to set them up. Examples are as follows:A merchant made a mixture of 150lb. of tea worth $109.50 by mixing tea worth $1.25 a pound with tea worth $.65 a pound. How many pounds of each kind did he use?
amount | price per lb | worth
-------------------------------------
x | $1.25 | 1.25x
-------------------------------------
y | $0.65 | 0.65y
-------------------------------------
amount | price per lb | worth
-------------------------------------
x | $1.25 | 1.25x
-------------------------------------
y | $0.65 | 0.65y
-------------------------------------
150 lb | ?? | $109.50
| salt | water | total
-------------------------------------
1st situation | 4 | 6 | 10
-------------------------------------
2nd situation | 4 | ? | ?
-------------------------------------
| salt | water | total
---------------------------------------
1st situation | 4 | 6 | 10
---------------------------------------
2nd situation | 4 | 6 + x | 10 + x
---------------------------------------
| tin | copper | total
--------------------! ------------------
| t | c | 110
----------! -------- ---------------------
