Find the x-coordinates of all relative minima of f(x). f(x)=(3)/(4)x^(4)+2x^(3)-36x^(2)+7 Answer: x=

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Find the  x-coordinates of all relative minima of  f(x).  f(x)=(3)/(4)x^(4)+2x^(3)-36x^(2)+7 Answer:  x=

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Find Derivative of f(x): 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. Let's start by finding the first derivative f'(x). f(x) = (3/4)x^4 + 2x^3 - 36x^2 + 7 Differentiate f(x) with respect to x: f'(x) = d/dx[(3/4)x^4] + d/dx[2x^3] - d/dx[36x^2] + d/dx[7] f'(x) = 4*(3/4)x^3 + 3*2x^2 - 2*36x + 0 f'(x) = 3x^3 + 6x^2 - 72xFind Critical Points: Now we need to find the critical points by setting f'(x) equal to zero and solving for x. 0 = 3x^3 + 6x^2 - 72x We can factor out a common factor of 3x: 0 = 3x(x^2 + 2x - 24) Now we factor the quadratic: 0 = 3x(x + 6)(x - 4) The critical points are x = 0, x = -6, and x = 4.Find Second Derivative: To determine which of these critical points are relative minima, we need to use the second derivative test. Let's find the second derivative f''(x). f''(x) = d/dx[f'(x)] f''(x) = d/dx[3x^3 + 6x^2 - 72x] f''(x) = 3*3x^2 + 2*6x - 72 f''(x) = 9x^2 + 12x - 72Evaluate Concavity at Critical Points: Now we evaluate the second derivative at each critical point to determine the concavity at those points. For x = 0: f''(0) = 9(0)^2 + 12(0) - 72 f''(0) = -72 Since f''(0) < 0, the function is concave down at x = 0, which means x = 0 is a relative maximum, not a minimum.Relative Minimum at x = -6: For x = -6: f''(-6) = 9(-6)^2 + 12(-6) - 72 f''(-6) = 9(36) - 72 - 72 f''(-6) = 324 - 72 - 72 f''(-6) = 180 Since f''(-6) > 0, the function is concave up at x = -6, which means x = -6 is a relative minimum.Relative Minimum at x = 4: For x = 4: f''(4) = 9(4)^2 + 12(4) - 72 f''(4) = 9(16) + 48 - 72 f''(4) = 144 + 48 - 72 f''(4) = 120 Since f''(4) > 0, the function is concave up at x = 4, which means x = 4 is also a relative minimum.

Find the $x$ -coordinates of all relative minima of $f(x)$ . $\newline$ $f(x)=\frac{3}{4} x^{4}+2 x^{3}-36 x^{2}+7$ $\newline$ 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)=34x4+2x3−36x2+7 f(x)=\frac{3}{4} x^{4}+2 x^{3}-36 x^{2}+7 f(x)=43​x4+2x3−36x2+7\newlineAnswer: x= x= x=

Find the $x$ -coordinates of all relative minima of $f(x)$ . $\newline$ $f(x)=\frac{3}{4} x^{4}+2 x^{3}-36 x^{2}+7$ $\newline$ Answer: $x=$