Find the relative extrema for the following functions by (1) finding the critical value(s) and (2)

determining whether at the critical value(s) the function is at a relative maximum or minimum.

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(a) f(x)=-9x + 126x-45 (1) Take the first derivative, set it equal to zero, and solve for x to find the critical value(s). f'(x)=-18x+126 = 0 x = 7 critical value (2) Take the second derivative, evaluate it at the critical value(s), and check for concavity to distinguish between a relative maximum and minimum. f"(x)=-18 f" (7)=-180 concave, relative maximum (b) f(x)=2x-18x + 48x29 (1) f'(x)=6x-36x + 48 = 0 f'(x)=6(x-6x+8)=0 f'(x)=6(x-2)(x-4) = 0 x = 2 x = 4 critical values (2) f" (x)=12x-36 - f" (2) 12(2) 36 = -12 < 0 concave, relative maximum convex, relative minimum f"(4) 12(4)- 36 = 12 > 0 (c) f(x) = x+8x380x + 195 (1) - f'(x) = 4x +24x - 160x = 0 f'(x) = 4x(x+6x-40) = 0 f'(x) = 4x(x-4)(x + 10) = 0 x = 0 x = 4 x = -10 critical values (2) "(x)=12x+48x-160 f"(-10)=12(-10) + 48(-10) - 160 = 560 > 0 f"(0) = 12(0)+48 (0) convex, relative minimum 160-160 < 0 concave, relative maximum = f" (4) 12(4)+48(4) 160 = 224 > 0 - convex, relative minimum