Answer:
The proof is detailed below.
Step-by-step explanation:
We will first prove that if H(x) is a differentiable function in [a,b] such that H'(x)=0 for all x∈[a, b] then H is constant. For this, take, x,y∈[a, b] with x<y. By the Mean Value Theorem, there exists some c∈(x,y) such that H(y)-H(x)=H'(c)(x-y). But H'(c)=0, thus H(y)-H(x)=0, that is, H(x)=H(y). Then H is a constant function, as it takes the same value in any two different points x,y.
Now for this exercise, consider H(x)=F(x)-G(x). Using differentiation rules, we have that H'(x)=(F-G)(x)'=F'(x)-G'(x)=0. Applying the previous result, F-G is a constant function, that is, there exists some constant C such that (F-G)(x)=C.
We have a three unknown, 4 equation homogeneous system. These always have at least (0,0,0) as a solution. Let's write the equations, one column at a time.
1a + 0b + 0c = 0
-1a + 1b +0c = 0
0a - 1b + c = 0
0a + 0b + -1 c = 0
We could do row reduction but these are easy enough not to bother.
Equation 1 says
a = 0
Equation 4 says
c = 0
Substituting in the two remaining,
-1(0) + 1b + 0c = 0
b = 0
0(0) - 1b + 0 = 0
b = 0
The only 3-tuple satisfying the vector equation is (a,b,c)=(0,0,0)
Use the power rule
d(x^n)/dx = n·x^(n-1)
to form the derivative.
f'(x) = 4·2x -4
f'(x) = 8x - 4
Now, substitute "a" for "x" to find f'(a):
f'(a) = 8a - 4
Answer:
7 i think not sure
Step-by-step explanation:
