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find the difference quotient of -x^2+9x

Find the difference quotient

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Mathway | Find the Difference Quotient f(x)=x^2+9x-7

Precalculus. Find the Difference Quotient f(x)=x^2+9x-7. ... [x 2 1 2 π ∫ ...
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Difference Quotient -

Difference Quotient. What is the difference quotient in calculus? We start with the definition and then we calculate the difference quotient for different functions.
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from the difference quotient that the elementary formulas for derivatives are developed. II.
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The Difference Quotient 3 |

find the difference quotient: Do the blob thang: * If you do ...
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algebra precalculus - Difference Quotient for $\;\frac 1{x^2 ...

... {1\over x^2}$$ we are asked to find the difference qu ... Difference Quotient for $\;\frac 1{x^2}$ ... $$ we are asked to find the difference quotient for $$ {g(x

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  • Mathway | Find the Difference Quotient f(x)=9x-x^3

    Precalculus. Find the Difference Quotient f(x)=9x-x^3. Consider the difference quotient formula.

    • difference quotient ?

      3 answers

      The difference quotient is [f(x+h) - f(x)]/h (your book may use some other notation, such as Δx, for the change in x) f(x+h) = -7(x+h)² + 9(x+h) - 6 Expanding this, f(x+h) = -7(x² + 2xh + h²) + 9(x+h) - 6 = -7x² - 14xh - 7h² + 9x + 9h - 6...

    • For the function f( x )=-7x^ 2 + 9x -6, find the difference quotient and...

      2 answers

      Recall that the difference quotient is: [f(x + h) - f(x)] / h So we have that the difference quotient is: f(x + h) = -7(x + h)^2 + 9(x + h) - 6 f(x) = -7x^2 + 9x - 6 [f(x + h) - f(x)] / h = [-7(x + h)^2 + 9(x + h) - 6 - (-7x^2 + 9x - 6)] /...

    • Finding the Difference Quotient !? (math)?

      1 answer

      By definition, the difference quotient is [f(x+h) - f(x)] / h For the given f(x), this is { [3(x+h)³ + 5(x+h) + 2] - (3x³ + 5x + 2) } / h If you expand the terms in the square brackets and then subtract the terms in parentheses, you should...
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  • Suggested Questions And Answer :

    f(x)=x^3 solve using the difference quotient

    To find the difference quotient at any point (x,f(x)), we need to find f(x) for a value of x just a tiny bit bigger than x. This value of x has a value of x+h, where h is a very tiny quantity. f(x+h)=(x+h)^3. This can be expanded as x^3+3x^2h+3xh^2+h^3. The difference between this and f(x)=x^3 is therefore 3x^2h+3xh^2+h^3. The difference in the x value is, of course, h by definition of h. So the difference quotient is (3x^2h+3xh^2+h^3)/h, being the vertical displacement divided by the horizontal displacement. This comes to 3x^2+3xh+h^2. Because h is very tiny, infinitesimally small, in fact, we can ignore the terms involving h, leaving us with 3x^2, the difference quotient. The difference quotient can be used to calculate approximations in evaluating the cubes of numbers close to known cubes. If we want the cube of 3.05, we know that 27 is the cube of 3, so we add 3*9*0.05=1.35 to 27 to get an approximation, i.e., 28.35. The actual value is closer to 28.373. 
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    How do I find the simplified form of the difference quotient for the function: f(x) = ax^4

    How do I find the simplified form of the difference quotient for the function: f(x) = ax^4  this is simple, remember that the a is a constant.  so there is NO PRODUCT. f'(x) = 4ax^3
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    Evaluate the difference quotient for the function f(x)=x^2+5

    To find the difference quotient consider f(x+h), where h is a very tiny value. If we evaluate this we get f(x+h)=(x+h)^2+5, which expands to x^2+2xh+h^2+5. If we subtract f(x)=x^2+5 from this we get 2xh+h^2. The difference quotient is this divided by h, giving us 2x+h. We ignore h because it's very tiny, infinitesimal. So the difference quotient is 2x. Note that the constant 5 disappears and the power of x is reduced.
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    With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

    Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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    Three divided by the sum of x and 3 equals the quotient of 9 and the difference of x and 3.

    ??????????????? "quoeshent" ??????????????... look tu me like yu hav 2 FRAKSHUNS... 3/(x+3) =9/(x-3)... 3=9*(x+3) /(x-3)... 3*(x-3) =9*(x+3)... 3x-9 =9x+27... (9x-3x) +(27+9)=0... 6x=36... x=36/6=6
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    find the difference quotient of -x^2+9x

    ???????????????????? "quoeshent" ?????????????????? ??? yu want FRAKSHUN ??? 9x-x^2 look like x*(9-x)
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    find the difference quotient and simplify answer -4y2+6y(1+0)-4y2+6y(1)/0

    estimate. then find the difference 607-568
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    a wire of length l is cut into two parts. One part is bent into a circle and the other into square

    If the radius of the circle is d then its area is (pi)d^2 and its circumference is 2(pi)d, the length of wire we need to make the circle. The length of the remainder of the wire is l-2(pi)d, out of which we make the square. So the side of the square is a quarter of this perimeter, 1/4(l-2(pi)d), and the area of the square is the square of this side, 1/16(l-2(pi)d)^2. The sum of the areas of the circle and square is (pi)d^2+1/16(l-2(pi)d)^2. We need the minimum value of this expression, where the variable is d. So we differentiate it with respect to d. That's the same as getting the difference quotient. The expansion of 1/16(l-2(pi)d)^2 is l^2/16-1/4(pi)ld+1/4(pi)^2d^2. [Please distinguish between 1 and l in the following.] The difference quotient is zero at a maximum or minimum, so we have 2(pi)d-l/4(pi)+1/2(pi)^2d=0. We can take out (pi), leaving 2d-l/4+1/2(pi)d=0. Multiply through by 4 to get rid of the fractions: 8d-l+2(pi)d=0, from which d=l/(8+2(pi)). Half the length of the side of the square is 1/8(l-2(pi)d). If we substitute for d in this expression we get 1/8(l-2(pi)l/(8+2(pi)))  = l/8((8+2(pi)-2(pi))/(8+2(pi)) = l/8(8/(8+2(pi)) = l/(8+2(pi)) = d (QED). Therefore the radius of the circle = half the length of the side of the square is either a maximum or minimum value of the expression for the sum of the areas of the circle and square. We can see that this expression gets bigger as d gets bigger, because (pi)d^2 has a positive value always, so we do indeed have a minimum rather than a maximum. We can substitute d=l/(2(4+(pi))) in the expression for the sum of the areas and we get the minimum: (pi)d^2+1/16(l-2(pi)d)^2 = (pi)l^2/4(4+(pi))^2 + 1/16(l-2(pi)l/(2(4+(pi)))^2 = (pi)l^2/4(4+(pi))^2 + l^2/16(1-(pi)/(4+(pi))^2 = (pi)l^2/4(4+(pi))^2 + l^2/16(4+(pi)-(pi))^2(4+(pi))^2 = (pi)l^2/4(4+(pi))^2 + l^2/(4+(pi))^2 = l^2/(4(4+(pi))) or (l^2/4)*1/(4+(pi)) Sorry, it was getting difficult to represent the expressions using this tablet so I've had to accelerate the last bit! I hope I didn't make any mistakes!
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    Find the difference quotient: 2x^2-x+3y=2y

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    if 43/240<x1/x2<17/94, x3 is minimum possible value of x2 find 53*x3

    Question: if 43/240 Read More: ...

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