If f(x) = 2x^6 - 3,then: | vedclass

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(C) To find the local extrema of the function $f(x) = 2x^6 - 3$,we first find its derivative.$f'(x) = \frac{d}{dx}(2x^6 - 3) = 12x^5$.Setting the deri

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$-3$ is the local minimum value of $f$

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(C) To find the local extrema of the function $f(x) = 2x^6 - 3$,we first find its derivative.$f'(x) = \frac{d}{dx}(2x^6 - 3) = 12x^5$.Setting the derivative equal to zero to find critical points:$12x^5 = 0 \implies x = 0$.Now,we use the first derivative test or second derivative test to check the nature of the point $x = 0$.Using the first derivative test:For $x For $x > 0$,$f'(x) = 12x^5 > 0$ (function is increasing).Since the derivative changes sign from negative to positive at $x = 0$,the function has a local minimum at $x = 0$.The local minimum value is $f(0) = 2(0)^6 - 3 = -3$.Thus,$-3$ is the local minimum value of $f$.

If $f(x) = 2x^6 - 3$,then:

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