Suppose f(x) is a differentiable function with f(-1)=2 and f(2)=-1. The differentiable function g(x) is defined by the formula g(x)=f(f(x))'
A. Compute g(-1) and g(2). Explain why g(x)=0 must have at least one solution A between -1 and 2.
B. Compute g'(-1) and g'(2) in terms of values of f and f'. Verify that g'(-1) = g'(2). Explain why g"(x)=0 must have at least one solution B between -1 and 2.
C. Suppose now that f(x)=Cx^2 +D. Find values of C and D so that f(-1)=2 and f(2)=-1. Compute g(x)=f(f(x)) directly for those values of C and D and use algebra on the resulting formulas for g(x) and g"(x) to find numbers A and B between -1 and 2 so that g(A)=0 and g"(B)=0. The "abstract" assertions of a) and b) should be verified.
A. To compute g(-1), we substitute x = -1 into the definition of g(x):
g(-1) = f(f(-1))
Since f(-1) = 2 (given), we have:
g(-1) = f(f(-1)) = f(2)
Similarly, to compute g(2), we substitute x = 2 into the definition of g(x):
g(2) = f(f(2))
Since f(2) = -1 (given), we have:
g(2) = f(f(2)) = f(-1)
To show that g(x) = 0 must have at least one solution A between -1 and 2, note that g(-1) and g(2) are evaluations of f(x) at two different points. Since f(x) is a continuous function (because it is differentiable), by the Intermediate Value Theorem, g(x) must take on all values between g(-1) and g(2) at some point between -1 and 2. Therefore, g(x) = 0 must have at least one solution A between -1 and 2.
B. To compute g'(-1) and g'(2), we take the derivative of g(x) with respect to x using the chain rule:
g'(x) = f'(f(x)) * f'(x)
Substituting x = -1, we have:
g'(-1) = f'(f(-1)) * f'(-1)
Since f(-1) = 2 and f'(-1) is the derivative of f(x) evaluated at x = -1, we can use the given information to compute g'(-1).
Similarly, substituting x = 2, we have:
g'(2) = f'(f(2)) * f'(2)
Since f(2) = -1 and f'(2) is the derivative of f(x) evaluated at x = 2, we can use the given information to compute g'(2).
To show that g'(-1) = g'(2), we need to verify that the expressions for g'(-1) and g'(2) are equal based on the given information.
To show that g"(x) = 0 must have at least one solution B between -1 and 2, we can differentiate g'(x) using the chain rule:
g"(x) = f''(f(x)) * [f'(x)]^2 + f'(f(x)) * f''(x)
Then, we can substitute x = B and set g"(B) = 0 to find the value of B, as requested.
C. In this part of the question, we are given f(x) = Cx^2 + D, and we need to find values of C and D that satisfy f(-1) = 2 and f(2) = -1.
Substituting x = -1 into the equation f(x) = Cx^2 + D, we have:
2 = C(-1)^2 + D
2 = C + D
Substituting x = 2 into the equation f(x) = Cx^2 + D, we have:
-1 = C(2)^2 + D
-1 = 4C + D
We now have a system of two equations with two unknowns (C and D):
2 = C + D
-1 = 4C + D
We can solve this system of equations to find the values of C and D.
Once we have found the values of C and D, we can compute g(x) directly using the formula g(x) = f(f(x)). Then, we can use algebra on the resulting formulas for g(x) and g"(x) to find the numbers A and B between -1 and 2 such that g(A) = 0 and g"(B) = 0, as requested.
A. To compute g(-1), we first need to find f(-1). Given that f(-1) = 2, we can substitute it into the formula g(x) = f(f(x))' to obtain g(-1) = f(f(-1))'.
Next, we need to find f(-1)'. Since f(x) is differentiable, its derivative exists. Therefore, we can find the derivative of f(x) and evaluate it at x = -1.
We are given that f(-1) = 2, so we know that the point (-1, 2) lies on the graph of f(x). To find f'(-1), we can use the formula for the derivative:
f'(-1) = lim(h->0) [f(-1 + h) - f(-1)] / h
Now, let's compute g(2) in a similar way. We first find f(2) = -1. Then, we calculate f'(2) using the formula for the derivative:
f'(2) = lim(h->0) [f(2 + h) - f(2)] / h
Now that we have obtained the values of f(2) and f'(2), we can substitute them into the formula g(x) = f(f(x))' to find g(2).
To show that g(x) = 0 must have at least one solution A between -1 and 2, we need to analyze the properties of f(x) and f'(x). Specifically, we need to determine if there exist values of x between -1 and 2 such that f(f(x)) = 0.
B. To compute g'(-1), we need to find the derivative of g(x) with respect to x and evaluate it at x = -1. We can find g'(x) using the chain rule:
g'(x) = f'(f(x)) * f'(x)
By substituting x = -1 into this equation, we get g'(-1) = f'(f(-1)) * f'(-1).
Similarly, to compute g'(2), we substitute x = 2 into the equation g'(x) = f'(f(x)) * f'(x), which gives us g'(2) = f'(f(2)) * f'(2).
To verify that g'(-1) = g'(2), we compare the values of f'(f(-1)) * f'(-1) and f'(f(2)) * f'(2).
Next, we need to explain why g"(x) = 0 must have at least one solution B between -1 and 2. To do this, we observe that g'(x) = f'(f(x)) * f'(x). Thus, g"(x) can be found by taking the derivative of g'(x) with respect to x. By applying the chain rule again, we find:
g"(x) = [f''(f(x)) * f'(x)] * f'(x) + f'(f(x)) * f''(x)
We can now analyze the behavior of g"(x) within the interval from -1 to 2 to determine if there exists a value B such that g"(B) = 0.
C. Let's suppose that f(x) = Cx^2 + D. We need to find the values of C and D that satisfy f(-1) = 2 and f(2) = -1.
Substituting these values into the equation, we have:
f(-1) = C(-1)^2 + D = C + D = 2
f(2) = C(2)^2 + D = 4C + D = -1
Now we have a system of two equations with two variables (C and D). We can solve this system to find the values of C and D that satisfy both equations.
After finding the values of C and D, we can compute g(x) = f(f(x)) using the given values of C and D. By simplifying the resulting formula for g(x), we can find the numbers A and B between -1 and 2 such that g(A) = 0 and g"(B) = 0, which allows us to verify the abstract assertions made in parts A and B.