f and g are inverse functions.
g(x) = 5. So whatever f(x) or f(-1) is, g of that is going to be 5.
f(x)=5x, g(x)=3x-7 gf(x)= = g(f(x)) = g(5x) = 3*(5x) - 7 = 15x-7 So gf(4) = 15*4-7=53
Conventionally, f o g is a way to represent the operation of function g followed by function f.So for example, suppose f(x) = x + 2 and g(x) = 3xthen f o g (x) = f[g(x)] = f(3x) = 3x + 2Note, though, that the "o" operator is not commutative.g o f (x) = g[f(x)] = g(x + 2) = 3*(x + 2) = 3x + 6 which is not the same as f o g(x).Such "nested" or "chained" functions are particularly useful in calculus (differentation and integration).
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)
To find (f-g)(x), we need to subtract g(x) from f(x). So, (f-g)(x) = f(x) - g(x). Substituting the given functions, we get (f-g)(x) = (x+8) - (-4x-3). Simplifying this expression, we get (f-g)(x) = x + 8 + 4x + 3 = 5x + 11. Therefore, (f-g)(x) = 5x + 11.
f(x)=x+1 g(f(x))=x f(x)-1=x g(x)=x-1
g(x) = 5. So whatever f(x) or f(-1) is, g of that is going to be 5.
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
Yes
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
Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
x = 75
x
gf(5) = g(f(5)) = g(5+1) since f(x) = x+1 and then g(6) = 3*6 - 2 = 18 - 2 = 16
f(x)=5x, g(x)=3x-7 gf(x)= = g(f(x)) = g(5x) = 3*(5x) - 7 = 15x-7 So gf(4) = 15*4-7=53
f(x) = x - 1g(x) = (x2 - 1)/(x + 1) = (x - 1 )(x + 1)/(x + 1) = x - 1therefore, f(x) = g(x)
Conventionally, f o g is a way to represent the operation of function g followed by function f.So for example, suppose f(x) = x + 2 and g(x) = 3xthen f o g (x) = f[g(x)] = f(3x) = 3x + 2Note, though, that the "o" operator is not commutative.g o f (x) = g[f(x)] = g(x + 2) = 3*(x + 2) = 3x + 6 which is not the same as f o g(x).Such "nested" or "chained" functions are particularly useful in calculus (differentation and integration).