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f(x) and g(x) are just names of generic functions - they could be anything. In any specific case, where they intersect depends on how the functions are defined. In general, to find out where they intersect you can solve for:
f(x) = g(x)
Replacing the corresponding expressions for each function of course.
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How would you describe the difference between the graphs of f(x)2x2 and g(x)-2x2?
If the two at the end of these are exponents, like x^2, then these graphs would be reflections across the x-axis. Their graphs would be two parabolas. f(x) pointing up, and g(x) pointing down.
Is -3x-6 greater than x-2?
That depends on the value of x. You can think of those two expressions as being linear functions: f(x) = -3x - 6 g(x) = x - 2 To find the point at which they intersect (where f(x) = g(x)), we simply have to declare them as being equal and solve for x: if f(x) = g(x) then: -3x - 6 = x - 2 -4x = 4 x = -1 So the two functions intersect at the point where x = -1. If you plug that value into either function, you'll find that they return the value -3. This tells ups that these functions describe two lines that intersect at the point (-1, -3). Given that the coefficient of our first function, "f" has a lower value than the function "g", we know that when the x value is greater than that of their point of intersection, "g" will return a greater value. Before our point of intersection, "f" will return a greater value. So in answer to your original question, we can say: if x < -1: yes if x ≥ -1: no
Which of the following will form the composite function gfx shown below gfx?
G(F(x)) =~F(x) = and G(x) = 1F(x) = + 1 and G(x) = 3xF(x) = x + 1 and G(x) =orF(x) = 3x and G(x) = + 1-F(x) = x+ 1 and G(x) =G(F(x)) = x4 + 3~F(x) = x and G(x) = x4F(x) = x + 3 and G(x) = x4F(x) = x4 and G(x) = 3orF(x) = x4 and G(x) = x+ 3-It's F(x) =x4 andG(x) = x+ 3G(F(x)) =4sqrt(x)F(x) = sqrt(x) and G(x) = 4x
what if f(x)=2x-6 and g(x)=3x+9 find (f-g)(x)?
If f(x)=2z^2+5 and g(x)=x^2-2, fine (f-g)(x)
what answer is this f(x)=3x+10 and g(x)=2x-4 find (f+g)(x)?
What answer is this f(x)=3x+10 and g(x)=2x-4 find (f+g)(x)?
What will the intersection of graphs of the two linear equations tell you?
y=f(x) and y =g(x) are two linear equation of x. the intersection of their graphs will tel the solution of the equation f(x)=g(x).
How would you describe the difference between the graphs of f(x)2x2 and g(x)-2x2?
If the two at the end of these are exponents, like x^2, then these graphs would be reflections across the x-axis. Their graphs would be two parabolas. f(x) pointing up, and g(x) pointing down.
Why does it make sense that by graphing two rational functions on the same graph be the same as if you were to add them first then graph?
It is not. If f(x) = ax2 and g(x) = -ax2 then the separate graphs will be two quadratic curves, f being "happy" and g being "sad". But f(x) + g(x) = 0 for all x and so is the x axis, not a quadratic.
How many solutions will a system of equations have if the graphs of the lines intersect?
A system of equations will have one solution if the graphs of the lines intersect. This is because the lines intersect at a single point. Let's say that point is (a, b). The x = a, y = b is the one and only solution for the system.
What is the integral of the quantity f prime times g minus f times g prime divided by the quantity f squared minus g squared with respect to x where f and g are functions of x?
∫ [f'(x)g(x) - f(x)g'(x)]/(f(x)2 - g(x)2) dx = (1/2)ln[(f(x) - g(x))/(f(x) + g(x))] + C
What is the derivative of f plus g of 3 and f times g of 3 given that f of 3 equals 5 d dx f of 3 equals 1.1 g of 3 equals -4 d dx g of 3 equals 7 Also please explain QUICK THANK YOU?
d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
Rules of differentiation?
Assume f=f(x), g=g(x)and (f^-1)(x) is the functional inverse of f(x). (f+g)'=f'+g' (f*g)'=f'*g+f*g' product rule (f(g))'=g'*f'(g) compositional rule (f/g)'=(f'*g-f*g')/(g^2) quotient rule (d/dx)(x^r)=r*x^(r-1) power rule and applies for ALL r. where g^2 is g*g not g(g)
If f of x equals x plus 1 and g of f of x equals x what is g of x?
f(x)=x+1 g(f(x))=x f(x)-1=x g(x)=x-1
What are the rules of differentiation?
While no set of rules can handle differentiating every expression, the following should help. For all of the following, assume c and n are constants, f(x) and g(x) are functions of x, and f'(x) and g'(x) mean the derivative of f and g respectively. Constant derivative rule:d/dx(c)=0 Constant multiple rule:d/dx(c*f(x))=c*f'(x) Sum and Difference Rule:d/dx(f(x)±g(x))=f'(x)±g'(x) Power rule:d/dx(xn)=n*xn-1 Product rule:d/dx(f(x)*g(x))=f'(x)*g(x) + g'(x)*f(x) Quotient rule:d/dx(f(x)/g(x))=(f'(x)*g(x)-g'(x)*f(x))/f(x)² Chain rule:d/dx(f(g(x))= f'(g(x))*g'(x)
What is the integral of the quantity f prime times g minus f times g prime divided by the quantity f squared plus g squared with respect to x where f and g are functions of x?
∫ [f'(x)g(x) - f(x)g'(x)]/(f(x)2 + g(x)2) dx = arctan(f(x)/g(x)) + C C is the constant of integration.
If f-1(x)g(x) inverse then the domain of g(x) the range of f(x)?
If f(x) is the inverse of g(x) then the domain of g(x) and the range of f(x) are the same.
What is shortest solve about fermat?
To: trantancuong21@yahoo.com PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
