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Find if any critical numbers of function f(x)=x^5-10x^3?

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Find if any critical numbers of function f(x)=x^5-10x^3?

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giorgiana1976's profile pic

Posted (Answer #1)

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To determine the critical values of a function, you'll have to calculate the 1st derivative of this function. The roots of the 1st derivative represent the critical values of the function.

We'll differentiate f(x) with respect to x:

f'(x) = 5x^4 - 30x^2

We'll set f'(x) = 0:

5x^4 - 30x^2 = 0

We'll divide by 5:

x^4 - 6x^2 = 0

We'll factorize by x^2;

x^2(x^2 - 6) = 0

We'll cancel each factor:

x^2 = 0

x1 = x2 = 0

x^2 - 6 = 0

x3 = sqrt6 and x4 = -sqrt6

The critical values of the given function are: {-sqrt6 ; 0 ; sqrt6}.

justaguide's profile pic

Posted (Answer #2)

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The critical points of a curve representing the function f(x) = x^5 - 10x^3 are those where the value of the function is maximum or minimum.

To determine these points we need to solve f'(x) = 0.

f'(x) = 5x^4 - 30x^2 = 0

=> x^2(x^2 - 6) = 0

=> x^2 = 0 and x^2 = 6

=> x = 0, x = -sqrt 6 and x = sqrt 6

The critical values of the function lie at x = 0, x = -sqrt 6 and x = sqrt 6

shaznl1's profile pic

Posted (Answer #3)

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Critical points means the maximum or minimum value of a function, where extends to mean that the slope of the function is 0.

If we are going to calculate the slope, we need to find the 1st derivative.

Using Power rule

f(x)=x^5-10x^3

f'(x)=5x^4-30x^2

If we want to find when the slope is 0, set the derivative to 0

5x^4-30x^2=0

factor the GCF out, which is 5x^2

5x^2(x^2-6)=0

gives us

5x^2=0

x=0

or

x^2-6=0

x^2=6

x=sqrt6

or

x=-sqrt6

The critical values of the function f(x)= x^5-10x^3 are

x=0 x= sqrt6 and x=-sqrt6

 

mfonda's profile pic

Posted (Answer #4)

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To find critical numbers, you can look for points where the first derivative is undefined or zero. So to start, let's find the first derivative:

`f(x) = x^5 - 10x^3 \implies f'(x) = 5x^4 - 30x^2 = 5x^2(x^2 - 6)`

Next we must find the zeros:

`5x^2(x^2 - 6) = 0 \implies x = 0, x = \sqrt(6), x = -\sqrt(6)`

We have now found the critical points and we see that there are three of them. To find the critcal numbers, we must evaulate f(x) at each of there points:

`c_1 = f(0) = 0^5 - 10(0)^3 = 0`

`c_2 = f(\sqrt{6}) = (\sqrt{6})^5 - 10(\sqrt{6})^3 = 36\sqrt{6} - 60\sqrt{6} = -24\sqrt{6}`

` ` `c_3 = f(-\sqrt{6}) = (-\sqrt{6})^5 - 10(-\sqrt{6})^3 = -36\sqrt{6} + 60\sqrt{6} = 24\sqrt{6}`

 

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