What Is The Inverse Of H X =13x^3

Let f be the function defined by f(x) =x^3 + x. If g(x) is the inverse of f(x) and g(2) =1, what is the value of the derivative of g at x=2?

Let’s do this somewhat systematically.We should define things in Leibniz notation first, for mathematical convenience:Let [math]y_1 = f(x) = x^3+x[/math], and let [math]y_2 = g(x) = f^{-1}(x)[/math]To find the inverse of a function, all we need to do is to reflect the function about the line [math]y=x[/math], which basically maps all points [math](x,y)[/math] to [math](y,x)[/math]. In effect, we are swapping the roles of [math]x[/math] and [math]y[/math] as independent and dependent variables. We can apply this to [math]y_1 = x^3+x[/math] to get:[math]x = {y_2}^3+y_2[/math]Hey, we have some information about [math]g’(x)[/math], so maybe it will help differentiating both sides of that equation (make sure to do the chain rule correctly!):[math]\frac{d}{dx}x = \frac{d}{dx}({y_2}^3+y_2)[/math][math]1 = 3{y_2}^2\frac{dy_2}{dx} + \frac{dy_2}{dx}[/math][math]1 = (3{y_2}^2+1)\frac{dy_2}{dx}[/math][math]\frac{dy_2}{dx} = \frac{1}{3{y_2}^2+1}[/math]This sort of manipulation is called implicit differentiation, and it really helps us here when we want the derivative of a function when we can’t (or is difficult to) solve a function in closed form.Now - funnily enough - we are almost done! It is now a matter of plugging and chugging. We are looking for [math]g’(2) = \frac{dy_2}{dx}(2)[/math], given that [math]g(2) = 1[/math]. So let’s revert back to the notation introduced in the problem:[math]g’(x)= \frac{1}{3{(g(x))}^2+1}[/math][math]g’(2)= \frac{1}{3{(g(2))}^2+1}[/math][math]g’(2)= \frac{1}{3{(1)}^2+1}[/math][math]g’(2)= \frac{1}{4}[/math]

Find the value of a and b so that x+1 and x-2 are the factors of x^4+ ax^3-3x^2+2x+b?

X=-1 and x=2X^4+ax^3–3x^2+2x+b=0Put x=-1(-1)^4+a(-1)^3–3(-1)^2+2(-1)+b=01-a-3–2+b=0-a+b-4 =0-a+b=4=>(1)Putx=216+8a-12+4+b=08a+b=-8=>(2)equ.(1)-(2)-9a=12a =-4/3Put equ.1 in a= -4/3-(-4/3)+b=44/3+b=4b=8/3

What are the maximum and minimum values of (x + 1/x)?

[math]f(x)=x+\dfrac{1}{x}[/math][math]\implies f’(x) = 1–\dfrac{1}{x^2}[/math]Critical point(s):[math]f’(x)=0[/math][math]1-\dfrac{1}{x^2}=0[/math][math]\implies x=\pm 1[/math][math]f’’(x)=\dfrac{2}{x^3}[/math][math]f’’(-1)=-2, f’’(1)=2[/math]Maximum value of [math]f(x): f(-1) = -2[/math]Minimum value of [math]f(x): f(1) = 2[/math]

If [math]x + y = 2 \text{ and } x^2 + y^2 = 2[/math], what is the value of [math]xy[/math] ? Could you break it down for me?

If it doesn't specify that x cannot equal y, than this can be resolved several waysFirst a system of equations,[math]x+y=2 [/math][math]x^2+y^2=2 [/math]Since there are two variables in both, we want to use one of the equations to isolate either x or y in terms of the other variable so that you obtain an equation y= or x=. For this example I shall use [math]x+y=2[/math] and subtract the x from both sides yielding [math]y=2-x[/math] (another way to write it is in linear form [math]y=-x+2[/math]Now you see we have an equation which gives us the value of y, in terms of x. This is important because by substituting the [math]-x+2[/math] (which is equivalent to y) into the second equation we get x^2+(-x+2)^2=2. Factoring out (-x+2)^2 → (x^2–4x+4) (but don't forget to add the rest) leaving you with the equation x^2+x^2–4x+4=2the first two x^2’s can be combined into 2x^2…—->(2x^2–4x+4)=2.to factor this out you need to subtract the 2 from both sides so your equation is set to 0, → 2x^2–4x+2=0. because the equation is set to zero you can divide the left side by 2 (since 0/2=0) thus getting x^2–2x+1=0factor that out into (x-1)(x-1) which means the only value for x can be one. Plug this back into the non-substituted equation x+y=2 |x=1 —-> 1+y=2, thus y=1. they must both be one. I thought this may have been a trick question and did it geometrically by graphing the circle x^2+y^2=2 and the line y=-x+2 and the point of intersection was tangental at exactly (1,1). If there is a specification that x and y have to have different value… 0.o then i can't even see how thats possible. However I am not versed in any advanced mathematics.The geometric representation is quite nice but I don't know how to present it on quora but your circle would be of radius sqrt[2] which happens to intersect the line y=-x+2 at exactly (1,1)

If f(x) is a polynomial such that f(1) =1, f(2) =2, f(3) =3 and f(4) =16. Find the value of f(5)?

Let us assume : [math]f(x) = ax^3 + bx^2 + cx + d [/math]Then we get, [math]f(1) = a + b + c + d = 1 ... (1)[/math][math]f(2) = 8a + 4b + 2c + d = 2 ... (2)[/math][math]f(3) = 27a + 9b + 3c + d = 3 ...(3)[/math][math]f (4) = 64a + 16b + 4c + d = 16 ...(4)[/math]Solving (1) and (2), We get[math]7a + 3b + c = 1 ...(5)[/math]Solving (1) and (3), We get[math]26a + 8b + 2c = 2 ...(6)[/math]Solving (1) and (4), We get[math]63a + 15b + 3c = 15 ...(7)[/math]Solving (5) and (6)[math]6a + b = 0[/math]Solving (5) and (7)[math]7a + b = 2[/math]This gives, a = 2, b = -12And then, c = 23 and d = - 12Thus,[math] f(x) = 2x^3 - 12x^2 + 23x - 12[/math]And So, [math]f(5) = 2*(5)^3 - 12*(5)^2 + 23(5) - 12 = 53[/math]The required answer is f(5) = 53

If a and b are the roots of the equation x2 +ax+b=0, then the least value of the expression is (x belongs to R)?

A2A.Given that a and b are roots of the equation x^2 +ax +b=0,we can apply the results for the sum and product of the roots.Sum of roots= a+b = -a,      Hence 2a+b=0.Product of roots= ab = b,Since if b becomes zero, both a and b will become zero. Hence in this case function will be f(x) = x^2. So it's minimum value will be 0.If a and b are non-zero, from above equation a=1.So b = -2.Hence function is f(x) = x^2 +x -2.Thus f'(x) = 2x+1=0, so x =-1/2.Putting this in f(x),f(-1/2) = 1/4 -1/2 -2 = -9/4.So minimum value in this case will be -9/4.