Math  /  Algebra

Question```latex Sketch a graph of the polynomial function f(x)=x3+4x2+3xf(x)=x^{3}+4 x^{2}+3 x. Use to complete the following: ff is \qquad on the intervals (,3)(-\infty,-3) and (1,0)(-1,0). ff is \qquad on the intervals (,2)(-\infty,-2) and (0.5,)(-0.5, \infty). ff is \qquad on the intervals (3,1)(-3,-1) and (0,)(0, \infty). ff is \qquad on the interval (2,0.5)(-2,-0.5). ```

Studdy Solution

STEP 1

What is this asking? We need to figure out whether our function f(x)f(x) is increasing or decreasing on certain intervals! Watch out! Don't mix up increasing and decreasing; think about the function's slope!

STEP 2

1. Find the derivative.
2. Find critical points.
3. Analyze intervals.

STEP 3

Let's **define** our function: f(x)=x3+4x2+3xf(x) = x^3 + 4x^2 + 3x.
To see where f(x)f(x) is increasing or decreasing, we need its derivative, f(x)f'(x).

STEP 4

Using the **power rule**, we get: f(x)=3x2+8x+3.f'(x) = 3x^2 + 8x + 3.

STEP 5

**Critical points** occur where f(x)=0f'(x) = 0 or f(x)f'(x) is undefined.
Our f(x)f'(x) is a polynomial, so it's defined everywhere.
We just need to solve f(x)=0f'(x) = 0: 3x2+8x+3=0.3x^2 + 8x + 3 = 0.

STEP 6

We can **factor** this quadratic equation as: (3x+1)(x+3)=0.(3x + 1)(x + 3) = 0. This gives us two **critical points**: x=3x = -3 and x=13x = -\frac{1}{3} (which is the same as 0.333...-0.333..., close to the 0.5-0.5 in the problem).

STEP 7

Our **critical points** divide the number line into intervals: (,3)(-\infty, -3), (3,13)(-3, -\frac{1}{3}), and (13,)(-\frac{1}{3}, \infty).
We'll test a point in each interval to see if f(x)f'(x) is positive (increasing) or negative (decreasing).

STEP 8

* For (,3)(-\infty, -3), let's try x=4x = -4.
Then f(4)=3(4)2+8(4)+3=4832+3=19>0f'(-4) = 3(-4)^2 + 8(-4) + 3 = 48 - 32 + 3 = 19 > 0.
So f(x)f(x) is **increasing** on this interval.

STEP 9

* For (3,13)(-3, -\frac{1}{3}), let's try x=1x = -1.
Then f(1)=3(1)2+8(1)+3=38+3=2<0f'(-1) = 3(-1)^2 + 8(-1) + 3 = 3 - 8 + 3 = -2 < 0.
So f(x)f(x) is **decreasing** on this interval.

STEP 10

* For (13,)(-\frac{1}{3}, \infty), let's try x=0x = 0.
Then f(0)=3(0)2+8(0)+3=3>0f'(0) = 3(0)^2 + 8(0) + 3 = 3 > 0.
So f(x)f(x) is **increasing** on this interval.

STEP 11

ff is **increasing** on the intervals (,3)(-\infty, -3) and (1,0)(-1, 0). ff is **increasing** on the intervals (,2)(-\infty, -2) and (0.5,)(-0.5, \infty). ff is **decreasing** on the intervals (3,1)(-3, -1) and (0,)(0, \infty). ff is **decreasing** on the interval (2,0.5)(-2, -0.5).

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