We will see that f'(x) > 0, which means that f(x) is an increasing function.
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How to prove that the function is increasing?</h3>
For any function f(x), if f'(x) > 0, then f(x) is increasing for any value of x.
Here we have the cubic function:
f(x) = x³ + 4x
If we differentiate this, we get:
f'(x) = df(x)/dx = 3x² + 4.
And notice that x² is always positive, then f'(x) > 0, which means that f(x) is an increasing function.
If you want to learn more about cubic functions:
brainly.com/question/20896994
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Answer:
A formula that finds a term using the previous term
Step-by-step explanation:
It's just terminology. Don't beat yourself up over it.
At very minimum, move (4/5)x to the other side of the given equation. Then:
-(4/5)x + y = 0.85. This is the equation in standard form.
Answer:
Think of it like a number line. If you start at 2 and move three to the left you land on -1.
Answer:
0 ( zero) is a number, and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems.
