Evaluating Functions Flashcards | Quizlet ~ Press News

f(x)=x^2. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music… Wolfram|Alpha brings expert-level knowledge andSolution for F(x)=10x+8 equation: Simplifying F(x) = 10x + 8 Multiply F * x xF = 10x + 8 Reorder the terms: xF = 8 + 10x Solving xF = 8 + 10x Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right.Let f (x) = x 3 + a x 2 + b x + 5 sin 2 x be an increasing fuction in the set of real numbers R. Then a & b satisfy the condition: Then a & b satisfy the condition: HardGraph f(x)=10-|x| Find the absolute value vertex. In this case, the vertex for is . Tap for more steps... To find the coordinate of the vertex, set the inside of the absolute value equal to . In this case, . Replace the variable with in the expression. Simplify . Tap for more steps...Divide f-2, the coefficient of the x term, by 2 to get \frac{f}{2}-1. Then add the square of \frac{f}{2}-1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

F(x)=10x+8 - solution

Simple and best practice solution for f(x)=10-x equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.The value of function f (x) at x=10 is 8.Ive got this function F(x) = 8/5x +4 to 5x-20/8 = y and would list it as f^-1 = 5x-20/8 what I not sure is can you reduce the -20 and 8. I know their is some rule with variables like this and some can't be touched. Please list the rule if you know it and answer the question. I will award the points quickly.13 >Evaluate the function for x = 10 , that is substitute 10 into any x in the function. f(10) = 10/2 + 8 = 5 + 8 = 13

If f(x) = xe^x(1 - x) , then f(x) is - Toppr Ask

f(x)=x-11 rArr f(4)=4-11=-7.5x+10=2x-11 One solution was found : x = -7 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :Make sure we get the Domain for f(x) right, Then also make sure that g(x) gets the correct Domain; Example: f(x) = √x and g(x) = x 2. The Domain of f(x) = √x is all non-negative Real Numbers. The Domain of g(x) = x 2 is all the Real Numbers. The composed function is: (gx^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x - 1 that's your answer. you can confirm this solution is correct by multiplying it by (x-1) and you will get x^8 - 1 back again. the start of your long division would look like this: x^8 + 0x^7 + 0x6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x^1 - 1 divided by (x-1) the start of your synthetic division would look likehope u like it :Dacapellas used:airplane 4 walls - f(x)instrumental used:pantomime - wsjnespero que goste :D

fx-x^2=-6x+8

Subtract x^2 from either side.

fx-x^2+6x=8

Add 6x to all sides.

fx-x^2+6x-8=0

Subtract Eight from each side.

\left(f+6\right)x-x^2-8=0

Combine all phrases containing x.

-x^2+\left(f+6\proper)x-8=0

All equations of the form ax^2+bx+c=Zero will also be solved the usage of the quadratic components: \frac-b±\sqrtb^2-4ac2a. The quadratic method provides two solutions, one when ± is addition and one when it is subtraction.

x=\frac-\left(f+6\proper)±\sqrt\left(f+6\proper)^2-4\left(-1\right)\left(-8\proper)2\left(-1\proper)

This equation is in same old shape: ax^2+bx+c=0. Substitute -1 for a, f+6 for b, and -8 for c in the quadratic formula, \frac-b±\sqrtb^2-4ac2a.

x=\frac-\left(f+6\proper)±\sqrt\left(f+6\right)^2+4\left(-8\proper)2\left(-1\right)

Multiply -4 instances -1.

x=\frac-\left(f+6\right)±\sqrt\left(f+6\right)^2-322\left(-1\right)

Multiply Four instances -8.

x=\frac-\left(f+6\right)±\sqrtf^2+12f+42\left(-1\proper)

Take the sq. root of \left(f+6\proper)^2-32.

x=\frac-\left(f+6\proper)±\sqrtf^2+12f+4-2

Multiply 2 times -1.

x=\frac\sqrtf^2+12f+4-f-6-2

Now remedy the equation x=\frac-\left(f+6\proper)±\sqrtf^2+12f+4-2 when ± is plus. Add -\left(f+6\right) to \sqrt12f+4+f^2.

x=-\frac\sqrtf^2+12f+42+\fracf2+3

Divide -f-6+\sqrt12f+4+f^2 by way of -2.

x=\frac-\sqrtf^2+12f+4-f-6-2

Now resolve the equation x=\frac-\left(f+6\right)±\sqrtf^2+12f+4-2 when ± is minus. Subtract \sqrt12f+4+f^2 from -\left(f+6\right).

x=\frac\sqrtf^2+12f+42+\fracf2+3

Divide -f-6-\sqrt12f+4+f^2 via -2.

x=-\frac\sqrtf^2+12f+42+\fracf2+3 x=\frac\sqrtf^2+12f+42+\fracf2+3

The equation is now solved.

fx-x^2=-6x+8

Subtract x^2 from either side.

fx-x^2+6x=8

Add 6x to either side.

\left(f+6\right)x-x^2=8

Combine all phrases containing x.

-x^2+\left(f+6\proper)x=8

Quadratic equations reminiscent of this one can be solved through completing the sq.. In order to complete the sq., the equation must first be in the shape x^2+bx=c.

\frac-x^2+\left(f+6\right)x-1=\frac8-1

Divide all sides by -1.

x^2+\fracf+6-1x=\frac8-1

Dividing via -1 undoes the multiplication via -1.

x^2+\left(-\left(f+6\right)\right)x=\frac8-1

Divide f+6 by -1.

x^2+\left(-\left(f+6\proper)\proper)x=-8

Divide 8 through -1.

x^2+\left(-\left(f+6\right)\right)x+\left(-\fracf2-3\proper)^2=-8+\left(-\fracf2-3\proper)^2

Divide -\left(f+6\proper), the coefficient of the x time period, via 2 to get -\fracf2-3. Then add the sq. of -\fracf2-Three to all sides of the equation. This step makes the left hand side of the equation a super square.

x^2+\left(-\left(f+6\proper)\proper)x+\frac\left(f+6\proper)^24=-8+\frac\left(f+6\proper)^24

Square -\fracf2-3.

x^2+\left(-\left(f+6\right)\proper)x+\frac\left(f+6\right)^24=\frac\left(f+6\proper)^24-8

Add -Eight to \frac\left(f+6\right)^24.

\left(x-\fracf2-3\proper)^2=\frac\left(f+6\proper)^24-8

Factor x^2+\left(-\left(f+6\proper)\right)x+\frac\left(f+6\right)^24. In common, when x^2+bx+c is a super sq., it may possibly all the time be factored as \left(x+\fracb2\right)^2.

\sqrt\left(x-\fracf2-3\right)^2=\sqrt\frac\left(f+6\proper)^24-8

Take the sq. root of both sides of the equation.

x-\fracf2-3=\frac\sqrtf^2+12f+42 x-\fracf2-3=-\frac\sqrtf^2+12f+42

Simplify.

x=\frac\sqrtf^2+12f+42+\fracf2+3 x=-\frac\sqrtf^2+12f+42+\fracf2+3

Subtract -\fracf2-Three from each side of the equation.