To find f(-3), we substitute -3 into the function f(x) = x^2 + x: f(-3) = (-3)^2 + (-3) = 9 - 3 = 6
If f(x) = 35/5 + 3 then its inverse is f(x) = 5/3*(x - 3).
Are you trying to solve for x? Fx = x2 - 3 x2 - Fx - 3 = 0 x2 - Fx = 3 x2 - Fx + (F/2)2 = 3 + (F/2)2 (x - F/2)2 = 3 + (F/2)2 x - F/2 = ±[ 3 + (F/2)2 ]1/2 x = F/2 ± [ 3 + (F/2)2 ]1/2
f(x)=x2-x3 f(2) = 4-8 = -4 f'(x) =2x-3x2 f'(2) = 4-12=-8 f''(x) = 2 -6x f''(2)= -10 f'''(x)= -6 f(n)(x) = 0 for all n > 3. f(x) = f(2) + (x-2) f'(2) / 1! + (x-2)2 f''(2) /2! + (x-2)3 f'''(2)/3! + . . . f(x) = -4 -8(x-2) -10(x-2)22/2-6(x-2)3/6 + 0 + 0 + ... f(x) = -4 -8(x-2) -5(x-2)2 - (x-2)3
f(x) = x2 + 3x - 2 then f'(x) = 2x + 3 and then then f'(2) = 2*2 + 3 = 4+3 = 7
To find f(-3), we substitute -3 into the function f(x) = x^2 + x: f(-3) = (-3)^2 + (-3) = 9 - 3 = 6
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
If f(x) = 35/5 + 3 then its inverse is f(x) = 5/3*(x - 3).
Even polynomial functions have f(x) = f(-x). For example, if f(x) = x^2, then f(-x) = (-x)^2 which is x^2. therefore it is even. Odd polynomial functions occur when f(x)= -f(x). For example, f(x) = x^3 + x f(-x) = (-x)^3 + (-x) f(-x) = -x^3 - x f(-x) = -(x^3 + x) Therefore, f(-x) = -f(x) It is odd
Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
It is f(x) = 4*3^x.
Given the function f(x) = 2x + 3 and a = -1, we can find f(a) as follows: f(a) = 2(-1) + 3 f(a) = -2 + 3 f(a) = 1 So, f(a) = 1. To graph f(x) and 1/f(x), we can plot several points and connect them to visualize the functions. Here are some points for f(x): For f(x): When x = -2, f(x) = 2(-2) + 3 = -1 When x = -1, f(x) = 2(-1) + 3 = 1 When x = 0, f(x) = 2(0) + 3 = 3 When x = 1, f(x) = 2(1) + 3 = 5 When x = 2, f(x) = 2(2) + 3 = 7 Now, to find 1/f(x), we take the reciprocal of each f(x) value: For 1/f(x): When x = -2, 1/f(x) = 1/(-1) = -1 When x = -1, 1/f(x) = 1/1 = 1 When x = 0, 1/f(x) = 1/3 ≈ 0.333 When x = 1, 1/f(x) = 1/5 ≈ 0.2 When x = 2, 1/f(x) = 1/7 ≈ 0.143 Now, we can plot these points and connect them to obtain the graphs of f(x) and 1/f(x).
3 x f x f x f x f x f x f = 3f6
When x = 7, f(x) = 3*(4-x) = 3*(-3) = -9
-1
Provided that the range of g(x) is the domain of f(x) then it is the composite function, called f of g of x.Note that f(g(x) ) is not the same as g(f(x).For example, if f(x) = x + 2 and g(x( = 3*x for real x, thenf(g(x)) = f(3*x) = 3*x + 2while g(f(x)) = g(x + 2) = 3*(x + 2) = 3*x + 6
By using the fundamental theorem of Calculus. i.e. The integral of f(x) = F(x), your limits are [a,b]. Solve: F(b) - F(a). The FTC, second part, says that if f is a continuous real valued function of [a,b] then the integral from a to b of f(x)= F(b) - F(a) where F is any antiderivative of f, that is, a function such that F'(x) = f(x). Example: Evaluate the integral form -2 to 3 of x^2. The integral form -2 to 3 of x^2 = F(-2) - F(3) = -2^3/3 - 3^3/3 = -8/3 - 27/3 = -35/3