Learn Unit III: Calculus with Free Lessons & Tips

To verify the Mean Value Theorem (MVT) for the function \( f(x) = x^2 + 2x + 3 \) on the interval \( [4, 6] \), we need to check three conditions:

1. \( f(x) \) is continuous on \( [4, 6] \).
2. \( f(x) \) is differentiable on \( (4, 6) \).
3. There exists a point \( c \) in \( (4, 6) \) such that \( f'(c) = \frac{{f(b) - f(a)}}{{b - a}} \), where \( a = 4 \) and \( b = 6 \).

Let's check these conditions:

1. **Continuity of \( f(x) \) on \( [4, 6] \)**:

The function \( f(x) = x^2 + 2x + 3 \) is a polynomial function and is continuous everywhere. Therefore, it is continuous on the interval \( [4, 6] \).

2. **Differentiability of \( f(x) \) on \( (4, 6) \)**:

The function \( f(x) = x^2 + 2x + 3 \) is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the interval \( (4, 6) \).

3. **Applying the MVT**:

We need to find \( f'(x) \) and then find a \( c \) such that \( f'(c) = \frac{{f(6) - f(4)}}{{6 - 4}} \).

\[ f'(x) = 2x + 2 \]

Now evaluate \( f'(x) \) at \( x = c \):

\[ f'(c) = 2c + 2 \]

Now evaluate \( f(6) \) and \( f(4) \):

At \( x = 6 \):
\[ f(6) = 6^2 + 2(6) + 3 = 36 + 12 + 3 = 51 \]

At \( x = 4 \):
\[ f(4) = 4^2 + 2(4) + 3 = 16 + 8 + 3 = 27 \]

Now apply MVT condition:

\[ f'(c) = \frac{{f(6) - f(4)}}{{6 - 4}} = \frac{{51 - 27}}{2} = \frac{24}{2} = 12 \]

Now, we need to find \( c \) such that \( 2c + 2 = 12 \):

\[ 2c + 2 = 12 \]
\[ 2c = 12 - 2 \]
\[ 2c = 10 \]
\[ c = 5 \]

So, there exists a point \( c \) in \( (4, 6) \) such that \( f'(c) = \frac{{f(6) - f(4)}}{{6 - 4}} \).

Therefore, the Mean Value Theorem is verified for the function \( f(x) = x^2 + 2x + 3 \) on the interval \( [4, 6] \), and \( c = 5 \) is the point that satisfies the theorem.

To find the general solution of the given differential equation:

dydx=1+y21+x2dxdy=1+1+x2y2

Let's rewrite the equation in a more suitable form:

dydx=11+y21+x2dxdy=11+1+x2y2

Now, this looks like a separable differential equation. We can rewrite it as:

dydx=11+y21+x2dxdy=11+1+x2y2 dydx=1+x2+y21+x2dxdy=1+x21+x2+y2

Now, we can separate variables:

1+x21+x2+y2dy=dx1+x2+y21+x2dy=dx

Integrating both sides:

∫1+x21+x2+y2 dy=∫dx∫1+x2+y21+x2dy=∫dx

Let's denote u=1+x2+y2u=1+x2+y2, then du=2y dydu=2ydy:

12∫1u du=x+C21∫u1du=x+C

12ln⁡∣u∣=x+C21ln∣u∣=x+C

Substitute back u=1+x2+y2u=1+x2+y2:

12ln⁡∣1+x2+y2∣=x+C21ln∣1+x2+y2∣=x+C

ln⁡∣1+x2+y2∣=2x+2Cln∣1+x2+y2∣=2x+2C

∣1+x2+y2∣=e2x+2C∣1+x2+y2∣=e2x+2C

1+x2+y2=Ae2x1+x2+y2=Ae2x

Where A=±e2CA=±e2C.

Finally, if we let A=e2CA=e2C, we can express the solution explicitly:

1+x2+y2=e2x1+x2+y2=e2x

y2=e2x−1−x2y2=e2x−1−x2

y=±e2x−1−x2y=±e2x−1−x2

So, the general solution of the given differential equation is:

y=±e2x−1−x2y=±e2x−1−x2

To find the integrating factor of the given differential equation (1−x2)dydx−xy=1(1−x2)dxdy−xy=1, we can use the formula for the integrating factor μ(x)μ(x), which is given by:

where P(x)P(x) is the coefficient of yy in the given differential equation.

In this case, P(x)=−xP(x)=−x.

So,

Therefore, the integrating factor μ(x)μ(x) is: