Question Find the range of $x$ values where the function $f(x) = x^3 + 3x^2 - 9x + 7$ is increasing.

Studdy Solution

STEP 1

Assumptions
1. The function is defined as $f(x)=x^{3}+3 x^{2}-9 x+7$ .
2. We are looking for the interval(s) where the function is increasing.

STEP 2

To determine where the function is increasing, we need to find the derivative of $f(x)$ with respect to $x$ , which gives us the slope of the tangent to the curve at any point.
$f'(x) = \frac{d}{dx}(x^{3}+3 x^{2}-9 x+7)$

STEP 3

Calculate the derivative of each term separately.
$f'(x) = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(3 x^{2}) - \frac{d}{dx}(9 x) + \frac{d}{dx}(7)$

STEP 4

Apply the power rule for differentiation to each term.
$f'(x) = 3x^{2} + 6x - 9 + 0$

STEP 5

Simplify the derivative.
$f'(x) = 3x^{2} + 6x - 9$

STEP 6

To find the intervals where the function is increasing, we need to determine where the derivative is positive. This requires finding the critical points where the derivative is zero or undefined.
$3x^{2} + 6x - 9 = 0$

STEP 7

Factor out the common factor of 3 from the derivative.
$3(x^{2} + 2x - 3) = 0$

STEP 8

Now we need to solve the quadratic equation $x^{2} + 2x - 3 = 0$ .

STEP 9

Factor the quadratic equation.
$(x + 3)(x - 1) = 0$

STEP 10

Set each factor equal to zero and solve for $x$ .
$x + 3 = 0 \quad \text{or} \quad x - 1 = 0$

STEP 11

Find the values of $x$ .
$x = -3 \quad \text{or} \quad x = 1$

STEP 12

The critical points divide the number line into intervals. We need to test each interval to see where the derivative is positive. The intervals are $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ .

STEP 13

Choose a test point from each interval and plug it into the derivative $f'(x)$ .
For the interval $(-\infty, -3)$ , choose $x = -4$ . For the interval $(-3, 1)$ , choose $x = 0$ . For the interval $(1, \infty)$ , choose $x = 2$ .

STEP 14

Evaluate $f'(x)$ at each test point.
For $x = -4$ : $f'(-4) = 3(-4)^{2} + 6(-4) - 9$ For $x = 0$ : $f'(0) = 3(0)^{2} + 6(0) - 9$ For $x = 2$ : $f'(2) = 3(2)^{2} + 6(2) - 9$

STEP 15

Calculate the value of the derivative at each test point.
For $x = -4$ : $f'(-4) = 3(16) - 24 - 9 = 48 - 24 - 9 = 15$ For $x = 0$ : $f'(0) = -9$ For $x = 2$ : $f'(2) = 3(4) + 12 - 9 = 12 + 12 - 9 = 15$

STEP 16

Determine the sign of the derivative at each test point.
For $x = -4$ : $f'(-4) > 0$ , so the function is increasing on $(-\infty, -3)$ . For $x = 0$ : $f'(0) < 0$ , so the function is decreasing on $(-3, 1)$ . For $x = 2$ : $f'(2) > 0$ , so the function is increasing on $(1, \infty)$ .

STEP 17

Combine the intervals where the function is increasing.
The function $f(x)$ is increasing on the intervals $(-\infty, -3)$ and $(1, \infty)$ .

STEP 18

Match the combined intervals with the given options.
The correct answer is (C) $x<-3$ or $x>1$ .

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