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Find the critical point(s) of the function ( f(x)={(6-12 x)}{6X), if any, leaving your answer(s) in exact form (no approximations). Enter DNE if there are no critical points.?

User Astm
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Final answer:

The critical point of the function f(x)=-72x^2+36x is found by taking the derivative, setting it equal to zero, and solving for x, which results in the single critical point at x = 1/4.

Step-by-step explanation:

To find the critical points of a function, we need to take the derivative of the function and set it equal to zero, solving for x. However, the function provided in the question seems incorrect: f(x)={(6-12x)}{6X}. The notation is unclear, but it probably means f(x)=(6-12x)(6x), which is a quadratic equation and can be written as f(x)=6x(6-12x) or f(x)=-72x^2+36x.



Let's find the derivative of f(x):

f'(x)=d/dx(-72x^2 + 36x) = -144x + 36.



We set the derivative equal to zero to find the critical points:

0 = -144x + 36.



Solving for x gives us x = 36/144, or simplifying, x = 1/4.



Therefore, the critical point is x = 1/4. This is the point where the derivative of the function is zero, which can correspond to a local maximum, a local minimum, or a saddle point.

User Lucca Mordente
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