For the function below, determine a. f(x-1) b. f(x+a)-f(x) and c. f(x^2)

Question

The function is 3√x +6(x=4)

Thanks in advanced.

Answer 1

With the given function just substitute the following for x: a) x-1, b) x+a (evaluate (f(x+a)-f(x), and c) x^2. Hence,

 

a) f(x-1) = 3√(x-1) +6(x=4) which implies f(4-1)=3√(4-1) +6 =3√3 +6

 

b) f(x+a)-f(x) = 3√(x+a) +6 -(3√x +6)= 3√(x+a) +6 -3√x -6 = 3√(x+a) - 3√x Note: the constants drop out, and √(x+a) - √x ≅ 3(√x +1/2*a/√x + … + √a) - 3√x = 3/2*a/√x + … + 3√a

 

Also, it is good practice to expand the radical √(x+a) because in calculus you're going to evaluate the expression above, divide it by a to get the slope of the line between points (x,f(x))and (x+a,f(x+a)), and then let a go to zero to get the slope of the line tangent to the point (x,f(x)). Moreover, when dividing the expression by a and letting it go to zero, the rest of the terms will vanish and you'll be left with 3/2/√x for the answer.

 

However, just evaluate f(4+a) - f(4) and leave it at that for this part of the problem. 

 

c) f(x^2) = 3√x^2 +6 = 3x +6 which implies f(4^2) = f(16) = 3*4 +6 = 12 +6 =18

 

Part b) I just wanted to give you a heads up in what problems you'll be getting in calculus since you are taking precalculus.