Question

How many relative extrema can the polynomial t(x)=3x2−5x+8x3
have?(1 point)
at most
R

Answer 1

To find the maximum number of relative extrema a polynomial can have, we can use the derivative of the polynomial. The given polynomial is:

\[
t(x) = 8x^3 + 3x^2 - 5x
\]

### Step 1: Find the first derivative

The first derivative \( t'(x) \) is calculated as follows:

\[
t'(x) = \frac{d}{dx}(8x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x)
\]

Calculating each term, we get:

\[
t'(x) = 24x^2 + 6x - 5
\]

### Step 2: Set the derivative to zero

To find the relative extrema, we set the first derivative equal to zero:

\[
24x^2 + 6x - 5 = 0
\]

### Step 3: Use the quadratic formula

We can use the quadratic formula to solve for \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 24 \), \( b = 6 \), and \( c = -5 \).

Calculating the discriminant:

\[
b^2 - 4ac = 6^2 - 4(24)(-5) = 36 + 480 = 516
\]

Since the discriminant is positive (\( 516 > 0 \)), there are two distinct real roots.

### Step 4: Determine relative extrema

Since \( t'(x) \) is a quadratic polynomial (degree 2), it can have at most 2 critical points (places where \( t'(x) = 0 \)), which could correspond to relative maxima or minima.

### Conclusion

Thus, the polynomial \( t(x) = 8x^3 + 3x^2 - 5x \) can have at most **2 relative extrema**.