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S_A_V [24]
4 years ago
14

What is the maximum number of possible extreme values for the function, F(x)=x^3-7x-6

Mathematics
2 answers:
Olin [163]4 years ago
5 0
2 is the max number because it's one less than the degree
notsponge [240]4 years ago
3 0

Answer:

The maximum number of possible extreme values for the function,

F(x)=x^3-7x-6 is:

2

Step-by-step explanation:

By the Theorem of extreme values of a polynomial function we have:

The graph of a  polynomial equation of degree n has atmost ( less than or equal to) "n-1" extreme values (  i.e. minima and/or maxima).

That means the total number of extreme values could be n-1, n-3, n-5 etc.

Hence, here we have a polynomial equation as:

F(x)=x^3-7x-6

i.e. we have a polynomial function of degree 3 i.e. n=3

So, the maximum number of possible extreme values that may exits is: 2

( Since n-1=3-1=2)

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I need a photo but ill give you some information this will be with exponents.

<em>Write down the problem. Here it is:</em>

<em>2^2(x+3) + 9 - 5 = 32</em>

<em>Resolve the exponent. Remember the order of operations: PEMDAS, which stands for Parentheses, Exponents, Multiplication/Division, and Addition/Subtraction. You can't resolve the parentheses first because x is in the parentheses, so you should start with the exponent, </em><em>2^2.2^2 = 4</em>

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<em>Do the addition and subtraction. Just add or subtract the remaining numbers. Here's how:</em>

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<em>4x + 16 - 16 = 32 - 16</em>

<em>4x = 16</em>

<em>Isolate the variable. To do this, just divide both sides of the equation by 4 to find x. </em><em>4x/4 = x </em><em>and </em><em>16/4 = 4</em><em>, so </em><em>x = 4.</em>

<em>4x/4 = 16/4</em>

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<em>Check your work. Just plug </em><em>x = 4</em><em> back into the original equation to make sure that it checks out. Here's how:</em>

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<em>2^2(4+3)+ 9 - 5 = 32</em>

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<em><u>Credit to wikiHow.com</u></em>

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