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]
F(x)=3x-5 determine whether the function is one-to-one. If it is, find a formula for its inverse.?
All linear equations are one-one/injections. Method I - f: R-->R In this case, you can substitue first few elements of R (Real numbers, domain) in the equation & you will observe that the image of f(1), f(2) etc is a real number. Hence, it exists in the range. Since you haven't been given the domain and the range of the function, you can prove it by method II. Method II - x1 =/ x2 (x1 & x2 € R) 3.x1 =/ 3.x2 3.x1-5 =/ 3.x2 -5 Therefore, f(x1) =/ f(x2) *Note : x1 is x subscript 1. =/ stands for 'not equal to'. It needs to be a bijective function, for its inverse to exist. For it to be a bijective function, it needs to be both onto & one-one. Onto function : For every y€R (co-domain of f), there exists an x €R(domain of f) such that f(x) =3x-5 Let f(x) = y Therefore, y= 3x-5 y=3x-5 (y+5)/3 = x Substitute x value ^ in f(x) f(x) = 3x-5 = 3[(y+5)/3] - 5 = (y+5) -5 [since, 3 & 3 gets cancelled) = y f(x) = y Therefore, f is onto. f is one-one & onto, therefore it is bijective. Hence, it 's inverse exists. Inverse: Let f(x) = y => x = f^(-1) (y) --- Read as f inverse of y. We know, x = (y+5)/3 (Derived above. Should be derived again here) Therefore, f^(-1) (y) = (y+5)/3 ( Since, f^(-1) (y) = x ) f^(-1) (x) = (x+5)/3 f inverse of x formula = (x+5)/3 Hope this helped.
Find the inverse of f(x) = (3x - 24)^4. Determine whether it is a function, and state its domain and range.?
y=(3x-24)^4 x=(3y-24)^4 x^.25=3y-24 (x^.25+24)/3=y yes its a function domain= x> or equal to 0 range= all real numbers
How do you determine whether a function has an inverse?
http://www.uncwil.edu/courses/mat111hb/f...
How do I find the inverse function of x^2-6x=f(x)?
There is no inverse function for x^2 - 6x = f(x), because the inverse of a square is not a function. It involves a positive or negative square root, which means there will be two possible values of f'(x) for each value of x, which violates the definition of a function and therefore f'(x) cannot be a function.
Find the inverse function of f(x) = 2x + 9?
Hi. This is how you do it x= 2(fx) + 9 x-9= 2(fx) x-9 ------ = (fx) 2 Thus: f^-1(x)= x-9 divided by 2.
If [math]f(x) = 7x + 12[/math], what is [math]{f}^{-1}(x)[/math] (the inverse function)?
Let f(x) = y. Now, to find the inverse function, we have to find value x in terms of y to find the inverse. So to answer your question y = 7x-12 Hence x= (y-12)/7 in the answer
Find the inverse function of f informally. Verify that f (f^ -1(x)) = x and f ^ -1(f(x)= x?
Having trouble with these problems.? Math I need help anyone be patient with me trying to understand? Find the inverse function of f informally. Verify that f (f^ -1(x)) = x and f ^ -1(f(x)= x 1. f(x) = 2x + 1 Show that f and g are inverse functions algebraically. 2. f(x) = x-9/ 4, g(x) = 4x + 9 Show that f and g are inverse fuctions algebraically Problems 3 & 4. 3. f(x) = x^3, g(x)= 3 sqaure root of x 4. f(x) = 1/1+x, x >_ 0; g(x) = 1 - x/ x, 0 < x < _ 1 Show that f and g are inverse functions (a) graphically and (b) numerically. 5. f (x) = 2x, g(x) = x/2 Determine algebraically whether the function is one-to-one. Verify your answer graphically Problems 6 & 7. 6. f(x)=3x + 5 7. f (x) = square root 2x+3 Find the inverse function of f algebraically. 8. f(x) = 3x 9. f(x) = square root 4 - x ^2, 0 _< x <_ 2 Last question Restrict the domain of the function f so that the function is one-to-one and has an inverse function. Then find the inverse function f ^ -1. State the domains and ranges of f and f^ -1. Explain your results. 10. f(x) = 1 - x ^4
How do you find inverse function of x + 1/x , given that f(x) is bigger or equal to 2?
Let the function y=x+(1/x). To find the inverse we need to make x the subject (write x as a function of y). Since the function should be greater than 2 we need to check the conditions for x.There are 2 possible graphs for this inverse function.