THIS VIDEO IS DESIGNED PRIMARILY FOR STUDENTS OF GRADE 12. IT GIVES A FAST AND STEP BY STEP APPROACH OF LAGRANGE'S MEAN VALUE THEOREM. SO WATCH THIS TODAY. AMAZING YOUTUBE VIDEO IS ALSO PROVIDED AT THE END.

2. PREVIEW:
EXPERIENCED MATH TEACHER
• EXPLAIN LAGRANGE’S MEAN VALUE THEOREM
• IT’S GEOMETRICAL INTERPRETATION
BONUS:
DISCUSS A FEW GRADE 12 QUESTION PAPERS

3. Here, I will only be stating Lagrange’s Mean Value Theorem and not discussing the proof
LAGRANGE’S MEAN VALUE THEOREM states that
If a function f is
i). Continuous on [a,b]
ii) f is derivable on (a,b)
Then there exists at least one real number c ∈ 𝑎, 𝑏 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑓′
𝑐 =
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎

4. Geometrical interpretation of Lagrange’s Mean Value Theorem
Lagrange’s Mean Value Theorem states that there exists at least one point
lying between A and B, the tangent at which is parallel to the chord AB.
B
A
B

5. A
B

6. Question 1 (GRADE XII MATH)
VERIFY Lagrange’s Mean Value Theorem for the function f(x) = 𝑥3
− 5𝑥2
− 3𝑥 𝑖𝑛 [1,3]
i)f being a polynomial function is continuous in [1,3]
ii) 𝑓′
𝑥 = 3 𝑥2
− 10 𝑥 − 3
f is derivable on (1, 3 )

7. By lagrange’s Mean Value Theorem, there exists at least one c ∈ 1,3 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑓′
𝑐 =
𝑓 3 − 𝑓(1)
3 − 1
3𝑐2 − 10𝑐 − 3 =
−27 + 7
2
3 𝑐2 − 10𝑐 − 3 = −10
3𝑐2 − 10𝑐 + 7 = 0

8. 3𝑐2
− 7𝑐 − 3𝑐 + 7 = 0
𝑐 3𝑐 − 7 − 3𝑐 − 7 = 0
3𝑐 − 7 𝑐 − 1 = 0
𝑐 =
7
3
, 1
𝑐 =
7
3
∈ (1,3)
Lagrange’s mean value theorem is satisfied

9. Question 2
Verify lagrange’s Mean Value Theorem for the function f(x) = sinx – sin 2x on [ 0,𝜋]
i) f being a trigonometric function is continuous on[ 0,𝜋]
ii) 𝑓′ 𝑥 = 𝑐𝑜𝑠𝑥 − 2 𝑐𝑜𝑠2𝑥
f is derivable on ( 0, 𝜋)
By lagrange’s theorem ∃ 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐 ∈ 0, 𝜋 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡

10. 𝑓′
𝑐 =
𝑓 𝜋 − 𝑓(0)
𝜋 − 0
cos c – 2 cos2c = 0
cos 𝑐 − 2 (2𝑐𝑜𝑠2 𝑐 − 1) = 0
cos 𝑐 − 4 𝑐𝑜𝑠2
𝑐 + 2 = 0

11. 4𝑐𝑜𝑠2 𝑐 − cos 𝑐 − 2 = 0
cos 𝑐 =
1 ± 33
8
cos c = 0.8431 or – 0.593
𝑐 = 32.530 𝑜𝑟 126.580 ∈ (0, 𝜋)
Hence lagrange’s Mean Value theorem satisfied

12. Question 3
Grade 12 math
Use lagrange’s mean value theorem to determine a point on the curve
𝑦 = 𝑥2 − 4 defined in [2,4], where the tangent is parallel to the chord joining
the endpoints of the curve
𝑓 𝑥 = 𝑥2 − 4
i) f is continuous on [2,4] being a polynomial function

13. 𝑓′
𝑥 =
𝑥
𝑥2 − 4
ii)
f is derivable on (2,4)
By lagrange’s Mean Value Theorem ∃ 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑐 ∈ 2,4 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑓′ 𝑐 =
𝑓 4 − 𝑓(2)
4 − 2 f(4 )= 12 = 2 3
f(2) = 0

14. 𝑐
𝑐2 − 4
=
2 3
2
𝑐2
= 3 (𝑐2
− 4)
𝑐2 = 3𝑐2 − 12
2𝑐2 = 12

15. 𝑐 = ± 6 = ±2.45
Lagrange’s mean value theorem is satisfied
𝑐 = 2.45 ∈ (2,4)
YOU are asked to determine the point at which the tangent is parallel to the chord
Students forget to do that.

16. 𝑦 = 𝑥2 − 4
𝑥 = 6
𝑦 = 2
𝑝𝑜𝑖𝑛𝑡 = ( 6, 2)

17. Question 4
Is lagrange’s Mean Value Theorem applicable for the function f(x) = 𝑥 𝑖𝑛 [−2,3]
𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑟𝑖𝑣𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 0
lagrange’s mean value theorem is not applicable
https://www.youtube.com/watch?v=RNAdeCD1ncw&list=PL26
R7TjUyi8zDE3-OcadJuTKlXApnFJG6
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