Mean Value Theorem: Meaning, Formula, Proof

Ans: Given function is f(x)=3x²+2x+5

Interval is [-1,1]

To find c 

f(x)=6x+2

f(-1)(a)=3×-1+2×-1+5=3-2+5

f(1)(b)=3×1+2×1+5=3+2+5

f(c)= \(\frac{f1-f(-1)}{1-(-1)}\)

6c+2=\(\frac{3+2+5-3-2+5}{2}\)

6c+2=\(\frac{4}{2}\)

6c+2=2

6c=2-2=0

c=0

Ans: If f(x) is defined and continuous on the interval [a, b] and differentiable on (a, b), the mean value theorem asserts that there is at least one number in c in the interval [a, b].

To find c

f(x)=2x

f(2)(a)=2×2=4

f(4)(b)=4×4=16

By using the formula 

f(c)=\(\frac{f(b)-f(a)}{b-a}\)

f(c)=\(\frac{16-4}{2}\)=\(\frac{12}{2}\)=6

2x=6

x=\(\frac{6}{2}\)

x=3

c=3

Ans: We were told in the problem description that the function (whatever it is) fits the requirements of the Mean Value Theorem so let’s start out this and enter in the known values.

f(0) – f(-7) = f' (c) (0-(-7)) → f(0) +3=7f'(c)

Solve for f(0)

f(0) = 7f'(c) -3

Finally, we'll look at what we know about derivatives. We're told that the derivative's maximum value is 2. As a result, entering in the greatest possible value of the derivative into f'(c above will give us the maximum value of in this situation. This results in:

f(0) =7f'(c) – 3 ≤7(2) – 3 = 11

So, the largest possible value is:

f(0) ≤ 11

Ans: f: [-5,5] →R is a differentiable 

We also know that every differentiable function is continuous. 

Therefore, f is differentiable and continuous both on (-5,5)

By mean value theorem, there exists some c in (-5,5)

Such that f'(c)) = \(\frac{f(b)-f(a)}{b-a}\)

Given that 

f'(x) does not vanish anywhere 

f'(x)≠0 for any value of x 

Thus, 

f'(c)≠0

\(\frac{f(5)-f(-5)}{5-(-5)}\) ≠ 0

\(\frac{f(5)-f(-5)}{5+5}\) ≠ 0

f(5) – f(-5) ≠ 0×10

f(5) – f(-5) ≠ 0

f(5) ≠ f(-5)

Hence proved 

Ans: We already know that f(x) has 2 roots. Let’s suppose they are a and b Assumingly, we now know that f(x) is continuous and differentiable everywhere, and this is especially true on [a,b] and distinguishable on (a,b).

f'(c)) = \(\frac{f(b)-f(a)}{b-a}\)

But we now need to recall that a and b are the roots of f(x)

f' (c) = \(\frac {0 - 0} {b - a} = 0\)

f(x) has a root at x=c

It's worth repeating that we don't know a value for c. We've merely demonstrated that it exists. We've also left out the fact that c is the sole root. More than one root is entirely possible.

Ans: Firstly we need to verify that Rolle’s theorem can be applied f(-2) = 0 and f(2) = 0 and f(x) is continuous and differentiable.

Then f(x) = 3x² - 4x

There must be one value of x between -2 and 2 for which f(x) = 0

3x² - 4=0

x² = \(\frac{4}{3}\)

x= + \(\frac{2\sqrt {3}}{3}\)

Hence there are two values which satisfy Rolle’s Theorem 

Ans: We are given the functionfx=x1+x and the interval 1,3. This function is continuous on the closed interval 1,3 since it is the quotient of continuous functions y=x and y=1+x.

\(f'(x) = \frac{(1+x)(1)-x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}\)

We can clearly see that f is differentiable on the open interval 1,3 .The Mean Value Theorem's assumptions have now been satisfied. Let's use the Mean Value Theorem to determine all potential c values in the open interval using the Mean Value Theorem (1,3). Then

Ans: The mean value theorem states that:

f has to satisfy the conditions given below

• On [a,b] f is continuous
• On [a,b] f is differentiable

Then for c ∈ (a,b)

\(f' (c) = \frac{f(b)-f(a)}{b-a}\)

The Rolle’s theorem states that:

• On [a,b] f is continuous
• On [a,b] f is differentiable
• f(a) = f(b)

Then for c ∈ (a,b)

f'(c) =0

Rolle's Theorem is a special case of the Mean Value Theorem, as may be seen. Rolle's Theorem, on the other hand, is fairly simple to establish.

Ans: First and foremost, the denominator in the Cauchy formula's left side is not zero:

Then by Rolle’s theorem 

There is a point (f(c),g(c)) on the curve where the tangent is parallel to the chord joining the ends and of the curve.

Ans: These functions' derivatives are: f'(x)=(x⁴)=4x³,g'(x)=(x²)=2x.

This number clearly falls inside the range of 1,2 and so fulfils the Cauchy theorem.