If Lp(t)=e4(At+B)+Ce5*Lp*​(*t*)=*e*4(*At*+*B*)+*Ce*5 denotes the best linear approximation of the function q(t)*q*(*t*) at the point t=3*t*=3, then find the value of A+B+C*A*+*B*+*C*.

To find the value of \(A + B + C\) in the best linear approximation \(Lp(t)\) of the function \(q(t)\) at the point \(t = 3\), we need to understand the form of the linear approximation and how it relates to the function \(q(t)\).

The best linear approximation of a function \(q(t)\) at a point \(t = a\) is given by:
\[ Lp(t) = q(a) + q'(a)(t - a) \]

Given the function \(q(t)\):
\[ q(t) = \begin{cases} 
(5t - 9)^{\frac{1}{t-2}} & \text{if } 0 \leq t < 2, \\
e^4 & \text{if } t = 2, \\
\frac{e^{t^2 - e^4}}{t - 2} & \text{if } t > 2,
\end{cases} \]

### Calculating \( q(3) \) and \( q'(3) \)

Since the given point is \( t = 3 \), we use the part of the function for \( t > 2 \):
\[ q(t) = \frac{e^{t^2 - e^4}}{t - 2} \]

1. **Value at \( t = 3 \)**:
\[ q(3) = \frac{e^{3^2 - e^4}}{3 - 2} = e^{9 - e^4} \]

2. **Derivative \( q'(t) \) for \( t > 2 \)**:
We need to differentiate \( q(t) = \frac{e^{t^2 - e^4}}{t - 2} \):

Using the quotient rule:
\[ q'(t) = \frac{(t - 2) \cdot \frac{d}{dt} (e^{t^2 - e^4}) - e^{t^2 - e^4} \cdot \frac{d}{dt}(t - 2)}{(t - 2)^2} \]
\[ q'(t) = \frac{(t - 2) \cdot e^{t^2 - e^4} \cdot \frac{d}{dt} (t^2 - e^4) - e^{t^2 - e^4}}{(t - 2)^2} \]
\[ q'(t) = \frac{(t - 2) \cdot e^{t^2 - e^4} \cdot 2t - e^{t^2 - e^4}}{(t - 2)^2} \]
\[ q'(t) = e^{t^2 - e^4} \cdot \frac{2t(t - 2) - 1}{(t - 2)^2} \]

At \( t = 3 \):
\[ q'(3) = e^{9 - e^4} \cdot \frac{2 \cdot 3 \cdot (3 - 2) - 1}{(3 - 2)^2} \]
\[ q'(3) = e^{9 - e^4} \cdot \frac{6 - 1}{1} \]
\[ q'(3) = 5e^{9 - e^4} \]

### Linear Approximation \(Lp(t)\)
The linear approximation \(Lp(t)\) around \(t = 3\) is:
\[ Lp(t) = q(3) + q'(3)(t - 3) \]
\[ Lp(t) = e^{9 - e^4} + 5e^{9 - e^4}(t - 3) \]

Given the form \( Lp(t) = e^4(At + B) + Ce^5 \):
Comparing this with \( Lp(t) = e^{9 - e^4} + 5e^{9 - e^4}(t - 3) \):

We note that \( e^4(At + B) \) and \( Ce^5 \) must match the terms of the linear approximation.

1. **Matching constant terms:**
\[ e^4 \cdot B + C \cdot e^5 = e^{9 - e^4} \]

2. **Matching linear terms:**
\[ e^4 \cdot A \cdot t = 5e^{9 - e^4} \cdot t \]
Since the coefficients must match, we get:
\[ e^4 \cdot A = 5e^{9 - e^4} \]
\[ A = 5e^{9 - e^4 - 4} \]

### Finding A + B + C

To find the specific values, we solve:
1. \( B \cdot e^4 + C \cdot e^5 = e^{9 - e^4} \)
2. \( A = 5e^{9 - e^4 - 4} \)

Thus:
\[ B = e^{9 - e^4 - 4} \]
\[ C = \frac{e^{9 - e^4 - 5}}{5} \]

Summing these:
\[ A + B + C = 5e^{9 - e^4 - 4} + e^{9 - e^4 - 4} + \frac{e^{9 - e^4 - 5}}{5} = 6 e^{9 - e^4 - 4} + \frac{e^{9 - e^4 - 5}}{5} = 7 e^{9 - e^4 - 4}\]
Therefore:
\[ A + B + C = 7 e^{9 - e^4 - 4}\] 

In conclusion:
\[ A + B + C = 7 e^{9 - e^4 - 4}\] 

Feel free to ask questions or need more details on this.
### Additional Questions
1. What is the formula for the best linear approximation of a function at a point?
2. How do you apply the quotient rule for differentiation?
3. What is the value of \( q(3) \) in terms of \( e \)?
4. What is the value of \( q'(3) \) in terms of \( e \)?
5. Why do we need the derivative to find the linear approximation?
6. What are the steps for matching the terms in a linear approximation?
7. How do you simplify the expression \( e^4(At + B) + Ce^5 \)?
8. How can you verify the continuity of \( q(t) \) at other points?

### Tip

When finding linear approximations, always compare coefficients of like terms to ensure consistency.

Explore how to find the values of A + B + C in a linear approximation problem involving exponential functions. This problem covers linear approximation, differentiation using the quotient rule, and application of exponential function properties.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation