Let P(x) `=a_(0)+a_(1)x^2+a_(2)x^(4)+cdots+a_(n)x^(2n)` with all `a_(i) gt0` , i=0,1,2,3, `cdots`,n . Then P(x) has

Options

A. neither maximum nor minimum
B. only one maximum
C. only one minimum
D. one maximum and one minimum

Answer

To solve the problem, we need to analyze the polynomial function given by:

\[ P(x) = a_0 + a_1 x^2 + a_2 x^4 + \cdots + a_n x^{2n} \]

where all coefficients \( a_i > 0 \) for \( i = 0, 1, 2, \ldots, n \).

### Step 1: Find the derivative \( P'(x) \)

To find the critical points, we first compute the derivative of \( P(x) \):

\[ P'(x) = 0 + 2a_1 x + 4a_2 x^3 + 6a_3 x^5 + \cdots + 2n a_n x^{2n-1} \]

This can be expressed as:

\[ P'(x) = 2a_1 x + 4a_2 x^3 + 6a_3 x^5 + \cdots + 2n a_n x^{2n-1} \]

### Step 2: Factor out \( x \)

We can factor \( x \) out of the derivative:

\[ P'(x) = x(2a_1 + 4a_2 x^2 + 6a_3 x^4 + \cdots + 2n a_n x^{2n-2}) \]

### Step 3: Analyze the critical points

Setting \( P'(x) = 0 \):

\[ x(2a_1 + 4a_2 x^2 + 6a_3 x^4 + \cdots + 2n a_n x^{2n-2}) = 0 \]

This gives us one critical point at \( x = 0 \).

### Step 4: Determine the behavior of \( P'(x) \)

Next, we need to analyze the sign of \( P'(x) \):

- For \( x > 0 \):
- All terms in \( P'(x) \) are positive since \( a_i > 0 \) and \( x^k > 0 \) for \( k \geq 0 \). Thus, \( P'(x) > 0 \).

- For \( x < 0 \):
- The term \( x \) is negative, but the polynomial \( 2a_1 + 4a_2 x^2 + 6a_3 x^4 + \cdots + 2n a_n x^{2n-2} \) is positive (since all coefficients are positive and \( x^2, x^4, \ldots \) are positive). Thus, \( P'(x) < 0 \).

### Step 5: Conclusion about the function \( P(x) \)

Since \( P'(x) \) changes from negative to positive at \( x = 0 \), we conclude that:

- \( P(x) \) has a minimum at \( x = 0 \).

### Final Result

Thus, the polynomial \( P(x) \) has one minimum at \( x = 0 \).