Suppose f and g are continuously∗ differentiable functions defined on [a, b] such that f0(x) = g0(x) for all x ∈ [a, b]. using the fundamental theorem of calculus, show that f and g differ by a constant. that is, show that there exists a c∈r such that f (x) g(x) = c.

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.

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove. 

an inaccurate proof leads to errors, or stating results that are not accurate.  which wil mess the probelm up you dont want this