`star` Rolle’s Theorem

`star` Mean Value Theorem

`star` Mean Value Theorem

`\color{green} ✍️` Let `f : [a, b] → R` be continuous on ` [a, b]` and differentiable on `(a, b)`,

such that `color{red}{f(a) = f(b)}`, where a and b are some real numbers.

`\color{green} ✍️` Then there exists some `c` in `(a, b)` such that `color{blue}{f ′(c) = 0 }.`

`\color{green} ✍️` In Fig, In each of the graphs, the slope becomes zero at least at one point.

That is precisely the claim of the Rolle’s theorem.

such that `color{red}{f(a) = f(b)}`, where a and b are some real numbers.

`\color{green} ✍️` Then there exists some `c` in `(a, b)` such that `color{blue}{f ′(c) = 0 }.`

`\color{green} ✍️` In Fig, In each of the graphs, the slope becomes zero at least at one point.

That is precisely the claim of the Rolle’s theorem.

Q 3155434364

Verify Rolle’s theorem for the function `y = x^2 + 2, a = – 2` and` b = 2` .

Class 12 Chapter 5 Example 42

Class 12 Chapter 5 Example 42

The function `y = x^2 + 2` is continuous in [– 2, 2] and differentiable in (– 2, 2).

Also f (– 2) = f ( 2) = 6 and hence the value of f (x) at – 2 and 2 coincide. Rolle’s

theorem states that there is a point c ∈ (– 2, 2), where f′ (c) = 0. Since f′ (x) = 2x, we

get c = 0. Thus at c = 0, we have f′ (c) = 0 and c = 0 ∈ (– 2, 2).

`\color{green} ✍️` Let `f : [a, b] → R` be a continuous function on `[a, b]` and differentiable on `(a, b)`.

`\color{green} ✍️` Then there exists some `c` in `(a, b)` such that

`color{red}{f'(c) =(f(b)-f(a))/(b-a)}`

`=>` Mean Value Theorem (MVT) is an extension of Rolle’s theorem.

`=>` In fig `y = f(x)` is given, We've already interpreted `f ′(c)` as the slope of the tangent to the curve `y = f (x)` at `(c, f (c))`.

`\color{green} ✍️` From the Fig `color{blue}{((f(b)-f(a))/(b-a))}` is the slope of the secant drawn between `(a, f (a))` and `(b, f (b))`.

`=>` `color{blue}{"The MVT states that there is a point" \ \ c \ \ "in" (a, b) "such that the slope of the tangent at"}`

`color{blue}{ (c, f(c)) "is same as the slope of the secant between" (a, f (a)) "and" (b, f (b))}`.

`\color{green} ✍️` In other words, there is a point `c` in `(a, b)` such that the tangent at `(c, f (c))` is parallel to the secant between `(a, f (a))` and `(b, f (b))`.

`\color{green} ✍️` Then there exists some `c` in `(a, b)` such that

`color{red}{f'(c) =(f(b)-f(a))/(b-a)}`

`=>` Mean Value Theorem (MVT) is an extension of Rolle’s theorem.

`=>` In fig `y = f(x)` is given, We've already interpreted `f ′(c)` as the slope of the tangent to the curve `y = f (x)` at `(c, f (c))`.

`\color{green} ✍️` From the Fig `color{blue}{((f(b)-f(a))/(b-a))}` is the slope of the secant drawn between `(a, f (a))` and `(b, f (b))`.

`=>` `color{blue}{"The MVT states that there is a point" \ \ c \ \ "in" (a, b) "such that the slope of the tangent at"}`

`color{blue}{ (c, f(c)) "is same as the slope of the secant between" (a, f (a)) "and" (b, f (b))}`.

`\color{green} ✍️` In other words, there is a point `c` in `(a, b)` such that the tangent at `(c, f (c))` is parallel to the secant between `(a, f (a))` and `(b, f (b))`.

Q 3175434366

Verify Mean Value Theorem for the function` f (x) = x^2` in the interval [2, 4].

Class 12 Chapter 5 Example 43

Class 12 Chapter 5 Example 43

The function `f (x) = x^2` is continuous in [2, 4] and differentiable in (2, 4) as its

derivative f′ (x) = 2x is defined in (2, 4).

Now, f (2) = 4 and f (4) = 16. Hence

` ( f(b) - f(a) )/(b-a) = (16-4)/(4-2) = 6`

MVT states that there is a point c ∈ (2, 4) such that f′ (c) = 6. But f′ (x) = 2x which

implies c = 3. Thus at c = 3 ∈ (2, 4), we have f′ (c) = 6.