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To find the coefficient of \( x^3 \) in the function \( h(x) = f(x+1) - g(x+2) \), we first need to expand each component.
Step 1: Expand \( f(x+1) \)
Starting with \( f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4 \), expand each term:
1. \( 10(x+1) = 10x + 10 \)
2. \( a_2(x+1)^2 = a_2(x^2 + 2x + 1) = a_2x^2 + 2a_2x + a_2 \)
3. \( a_3(x+1)^3 = a_3(x^3 + 3x^2 + 3x + 1) = a_3x^3 + 3a_3x^2 + 3a_3x + a_3 \)
4. \( (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 \)
Collecting terms for the \( x^3 \) coefficient in \( f(x+1) \):
\( a_3x^3 + 4x^3 = (a_3 + 4)x^3 \)
Step 2: Expand \( g(x+2) \)
Next, expand \( g(x+2) = b_1 + 3(x+2) + b_2(x+2)^2 + b_3(x+2)^3 + (x+2)^4 \):
1. \( 3(x+2) = 3x + 6 \)
2. \( b_2(x+2)^2 = b_2(x^2 + 4x + 4) = b_2x^2 + 4b_2x + 4b_2 \)
3. \( b_3(x+2)^3 = b_3(x^3 + 6x^2 + 12x + 8) = b_3x^3 + 6b_3x^2 + 12b_3x + 8b_3 \)
4. \( (x+2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16 \)
Collecting terms for the \( x^3 \) coefficient in \( g(x+2) \):
\( b_3x^3 + 8x^3 = (b_3 + 8)x^3 \)
Step 3: Calculate the coefficient of \( x^3 \) in \( h(x) \)
The coefficient of \( x^3 \) in \( h(x) = f(x+1) - g(x+2) \) is:
\((a_3 + 4) - (b_3 + 8) = a_3 + 4 - b_3 - 8 = a_3 - b_3 - 4\)
Step 4: Conclusion
Since \( f(x) \neq g(x) \) for any \( x \), it implies \( a_3 \neq b_3 \), the specific values don't make a difference to the meaning of this expression, the coefficient remains constant as:-4
Let \(f(x)=x+log_{e}x−xlog_{e}x,\text{ }x∈(0,∞)\).