Find the x-coordinates of all relative minima of f(x). f(x)=2x^(3)-15x^(2)+17 Answer: x=

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Find the  x-coordinates of all relative minima of  f(x).  f(x)=2x^(3)-15x^(2)+17 Answer:  x=

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Find Derivative: To find the relative minima of the function f(x), we need to find the critical points by taking the derivative of f(x) and setting it equal to zero. f(x) = 2x^3 - 15x^2 + 17 Let's find the first derivative f'(x). f'(x) = d/dx (2x^3 - 15x^2 + 17) f'(x) = 6x^2 - 30xFind Critical Points: Now we need to find the critical points by setting the derivative equal to zero and solving for x. 0 = 6x^2 - 30x 0 = 6x(x - 5) This gives us two critical points: x = 0 and x = 5.Second Derivative Test: To determine whether these critical points are relative minima, we need to use the second derivative test or the first derivative test. Let's find the second derivative of f(x). f''(x) = d/dx (6x^2 - 30x) f''(x) = 12x - 30Evaluate at Critical Points: Now we will evaluate the second derivative at the critical points to determine if they are relative minima. For x = 0: f''(0) = 12(0) - 30 f''(0) = -30 Since f''(0) < 0, the function is concave down at x = 0, which means x = 0 is not a relative minimum.Conclusion: For x = 5: f''(5) = 12(5) - 30 f''(5) = 60 - 30 f''(5) = 30 Since f''(5) > 0, the function is concave up at x = 5, which means x = 5 is a relative minimum.

Find the $x$ -coordinates of all relative minima of $f(x)$ . $\newline$ $f(x)=2 x^{3}-15 x^{2}+17$ $\newline$ Answer: $x=$

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Find the x x x-coordinates of all relative minima of f(x) f(x) f(x).\newlinef(x)=2x3−15x2+17 f(x)=2 x^{3}-15 x^{2}+17 f(x)=2x3−15x2+17\newlineAnswer: x= x= x=

Find the $x$ -coordinates of all relative minima of $f(x)$ . $\newline$ $f(x)=2 x^{3}-15 x^{2}+17$ $\newline$ Answer: $x=$