Let P(x) be a polynomial of degree 5 having extremum at `x= -1 , 1 and underset(x rarr 0)(lim) ((P (x))/(x^(3)) -1)= 7`. The value of |P(7)| is

Answer

To solve the problem step by step, we will analyze the given information and derive the polynomial \( P(x) \) of degree 5.

### Step 1: Understanding the Polynomial
Given that \( P(x) \) is a polynomial of degree 5 with extremum points at \( x = -1 \) and \( x = 1 \), we can express \( P(x) \) in the general form:
\[ P(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f \]

### Step 2: Analyzing the Limit
We have the limit condition:
\[ \lim_{x \to 0} \left( \frac{P(x)}{x^3} - 1 \right) = 7 \]
This can be rewritten as:
\[ \lim_{x \to 0} \frac{P(x)}{x^3} = 8 \]
This implies that as \( x \to 0 \), \( P(x) \) behaves like \( 8x^3 \).

### Step 3: Setting Up the Polynomial
Since \( P(x) \) is a degree 5 polynomial, we can express it as:
\[ P(x) = ax^5 + bx^4 + 8x^3 + dx^2 + ex + f \]
For the limit to hold, the terms \( ax^5 \) and \( bx^4 \) must vanish as \( x \to 0 \). Therefore, we can write:
\[ P(x) = ax^5 + bx^4 + 8x^3 \]

### Step 4: Finding the Derivative
To find the extremum points, we need to differentiate \( P(x) \):
\[ P'(x) = 5ax^4 + 4bx^3 + 24x^2 \]
Setting \( P'(-1) = 0 \) and \( P'(1) = 0 \) gives us two equations.

### Step 5: Setting Up the Equations
1. For \( x = 1 \):
\[ 5a + 4b + 24 = 0 \quad \text{(Equation 1)} \]
2. For \( x = -1 \):
\[ 5a - 4b + 24 = 0 \quad \text{(Equation 2)} \]

### Step 6: Solving the Equations
Adding Equation 1 and Equation 2:
\[ (5a + 4b + 24) + (5a - 4b + 24) = 0 \implies 10a + 48 = 0 \implies 10a = -48 \implies a = -\frac{24}{5} \]
Substituting \( a \) back into Equation 1:
\[ 5\left(-\frac{24}{5}\right) + 4b + 24 = 0 \implies -24 + 4b + 24 = 0 \implies 4b = 0 \implies b = 0 \]

### Step 7: Final Form of the Polynomial
Now substituting \( a \) and \( b \) back into \( P(x) \):
\[ P(x) = -\frac{24}{5}x^5 + 8x^3 \]

### Step 8: Finding \( P(7) \)
Now we calculate \( P(7) \):
\[ P(7) = -\frac{24}{5}(7^5) + 8(7^3) \]
Calculating \( 7^5 \) and \( 7^3 \):
\[ 7^5 = 16807, \quad 7^3 = 343 \]
Thus,
\[ P(7) = -\frac{24}{5}(16807) + 8(343) = -\frac{403368}{5} + 2744 \]
Converting \( 2744 \) to a fraction:
\[ 2744 = \frac{13720}{5} \]
Now combining:
\[ P(7) = -\frac{403368}{5} + \frac{13720}{5} = -\frac{389648}{5} \]

### Step 9: Finding the Absolute Value
Finally, we need the absolute value:
\[ |P(7)| = \left| -\frac{389648}{5} \right| = \frac{389648}{5} = 77929.6 \]

### Final Answer
Thus, the value of \( |P(7)| \) is:
\[ \boxed{77929.6} \]