How to Verify Lagrange`s Mean Value Theorem

Practice Questions With Step-by-Step Solutions

Question 1

Verify LMVT for [f(x)=x^{2}-2x] on [[0,3]].

Step-by-Step Solution:

1. Polynomial → continuous & differentiable ✔
2. Average slope:
   [\dfrac{f(3)-f(0)}{3-0}][=\dfrac{3^{2}-6-0}{3}][=\dfrac{3}{3}=1]
3. Derivative:
   [f'(x)=2x-2]
4. Solve:
   [2c-2=1 ⇒ c=\dfrac{3}{2}]

Conclusion: LMVT holds at [c=\dfrac{3}{2}].

Question 2

Apply LMVT to [f(x)=\sin x] from [[0,\pi]].

Solution:

1. Continuous and differentiable ✔
2. Average slope:
   [\left(\dfrac{0-0}{\pi}=0\right)]
3. Derivative: [\cos c=0]
4. [c=\dfrac{\pi}{2}]

Question 3

Apply LMVT to [f(x)=\ln x] on [[1,e]].

Solution:

1. Continuous & differentiable ✔
2. Average slope:
   [\dfrac{1-0}{e-1}][=\dfrac{1}{e-1}]
3. Derivative:
   [\left(\dfrac{1}{c}=\dfrac{1}{e-1} ⇒ c=e-1\right)]

Question 4

Verify LMVT for [f(x)=x^{3}] on [[-1,2]].

Solution:

1. Polynomial ✔
2. Average slope:
   [\dfrac{8-(-1)}{3}=\dfrac{9}{3}=3]
3. Derivative: [3c^{2}=3 ⇒ c=±1]
4. Valid inside [(-1,2)]: [c=1]

Question 5

Apply LMVT to [f(x)=2\cos x] on [[0,\pi]].

Solution:

1. Trig → OK ✔
2. Average slope:
   [\dfrac{-2 – 2}{\pi}= \dfrac{-4}{\pi}]
3. [
   \dfrac{d}{dx}(2\cos x)=-2\sin x
   ]
4. Solve:
   [-2\sin c=\dfrac{-4}{\pi} ⇒ \sin c=\dfrac{2}{\pi}]
   A valid c exists in [(0,π)].

Question 6

Apply LMVT to [f(x)=\sqrt{x}] on [[1,4]].

Solution:

1. Continuous and differentiable ✔
2. Average slope:
   [
   \dfrac{2-1}{3}=\dfrac{1}{3}
   ]
3. Derivative:
   [
   [f'(x)=\dfrac{1}{2\sqrt{x}}]
   ]
4. Solve:
   [\dfrac{1}{2\sqrt{c}}][=\dfrac{1}{3} ][⇒ \sqrt{c}=\dfrac{3}{2} ][⇒ c=\dfrac{9}{4}]

Question 7

Test LMVT for [f(x)=\tan x] on [[0,\dfrac{\pi}{4}]].

Solution:

1. Continuous & differentiable ✔
2. Average slope:
   [
   \dfrac{1-0}{\pi/4}= \dfrac{4}{\pi}
   ]
3. Derivative:
   [
   [f'(x)=\sec^{2}x]
   ]
4. [
   \sec^{2}c=\dfrac{4}{\pi}
   ]
   → valid c exists.

Question 8

Apply LMVT to [f(x)=x\ln x] on [[1,e]].

Solution:

1. Continuous & differentiable ✔
2. Average slope:
   [
   \dfrac{e\cdot1 – 0}{e-1}=\dfrac{e}{e-1}
   ]
3. Derivative:
   [
   [f'(x)=\ln x + 1]
   ]
4. Solve:
   [
   \ln c + 1 = \dfrac{e}{e-1}
   ]
   → c exists.

Question 9

Verify LMVT for [f(x)=x^{2}+1] on [[-2,1]].

Solution:

1. Polynomial ✔
2. Average slope:
   [
   \dfrac{2-5}{3}=\dfrac{-3}{3}=-1
   ]
3. Derivative:
   [
   [f'(x)=2x]
   ]
4. Solve:
   [
   2c=-1 ⇒ c=-\dfrac{1}{2}
   ]

Question 10

Apply LMVT for [f(x)=e^{x}] on [[0,2]].

Solution:

1. Continuous & differentiable ✔
2. Average slope:
   [
   \dfrac{e^{2}-1}{2}
   ]
3. Derivative:
   [
   [f'(c)=e^{c}]
   ]
4. Solve:
   [e^{c} = \dfrac{e^{2}-1}{2}][ ⇒ c = \ln\left(\dfrac{e^{2}-1}{2}\right)]