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Math  /  Algebra

QuestionDecide whether the following statement is true or false. The cube function is odd and is increasing on the interval (−∞,∞)(-\infty, \infty).
Choose the correct answer below. True False

Studdy Solution

STEP 1

1. A function f(x) f(x) is odd if f(−x)=−f(x) f(-x) = -f(x) for all x x in the domain.
2. A function is increasing on an interval if for any x1<x2 x_1 < x_2 , f(x1)<f(x2) f(x_1) < f(x_2) .

STEP 2

1. Verify if the cube function is odd.
2. Verify if the cube function is increasing on the interval (−∞,∞)(- \infty, \infty).
3. Decide if the statement is true or false based on the verifications.

STEP 3

The cube function is defined as f(x)=x3 f(x) = x^3 . To check if it is odd, we need to verify if f(−x)=−f(x) f(-x) = -f(x) .
Calculate f(−x) f(-x) :
f(−x)=(−x)3=−x3 f(-x) = (-x)^3 = -x^3
Since f(−x)=−x3 f(-x) = -x^3 and −f(x)=−x3 -f(x) = -x^3 , we have f(−x)=−f(x) f(-x) = -f(x) .
Thus, the cube function is odd.

STEP 4

To verify if the cube function is increasing on the interval (−∞,∞)(- \infty, \infty), consider the derivative of the function f(x)=x3 f(x) = x^3 .
Calculate the derivative f′(x) f'(x) :
f′(x)=3x2 f'(x) = 3x^2
Since 3x2≥0 3x^2 \geq 0 for all x x and f′(x)=0 f'(x) = 0 only at x=0 x = 0 , the function is non-decreasing everywhere and strictly increasing where x≠0 x \neq 0 .
Thus, the cube function is increasing on the interval (−∞,∞)(- \infty, \infty).

STEP 5

Since both conditions are satisfied (the cube function is odd and increasing on (−∞,∞)(- \infty, \infty)), the statement is true.
The correct answer is: True\boxed{\text{True}}.

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