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Let f(x)=-4x^(3)+6x^(2)+1. What is the absolute minimum value of f over the closed interval -4 <= x <= 3 ? Choose 1 answer: (A) -53 (B) -353 (C) -78 (D) 3

Let f(x)=−4x3+6x2+1 f(x)=-4 x^{3}+6 x^{2}+1 .\newlineWhat is the absolute minimum value of f f over the closed interval −4≤x≤3 -4 \leq x \leq 3 ?\newlineChoose 11 answer:\newline(A) −53-53\newline(B) −353-353\newline(C) −78-78\newline(D) 33

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Q. Let f(x)=−4x3+6x2+1 f(x)=-4 x^{3}+6 x^{2}+1 .\newlineWhat is the absolute minimum value of f f over the closed interval −4≤x≤3 -4 \leq x \leq 3 ?\newlineChoose 11 answer:\newline(A) −53-53\newline(B) −353-353\newline(C) −78-78\newline(D) 33
  1. Find Derivative: To find the absolute minimum value of the function on the closed interval, we need to find the critical points of the function within the interval and evaluate the function at those points and at the endpoints of the interval.\newlineFirst, we find the derivative of the function f(x)=−4x3+6x2+1f(x) = -4x^3 + 6x^2 + 1.\newlinef′(x)=ddx(−4x3+6x2+1)=−12x2+12xf'(x) = \frac{d}{dx} (-4x^3 + 6x^2 + 1) = -12x^2 + 12x
  2. Find Critical Points: Next, we set the derivative equal to zero to find the critical points.\newline−12x2+12x=0-12x^2 + 12x = 0\newline12x(−x+1)=012x(-x + 1) = 0\newlineThis gives us two critical points: x=0x = 0 and x=1x = 1.
  3. Evaluate Function: Now we need to evaluate the function f(x)f(x) at the critical points and at the endpoints of the interval, which are x=−4x = -4 and x=3x = 3.
    f(−4)=−4(−4)3+6(−4)2+1=−4(−64)+6(16)+1=256+96+1=353f(-4) = -4(-4)^3 + 6(-4)^2 + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353
    f(0)=−4(0)3+6(0)2+1=1f(0) = -4(0)^3 + 6(0)^2 + 1 = 1
    f(1)=−4(1)3+6(1)2+1=−4+6+1=3f(1) = -4(1)^3 + 6(1)^2 + 1 = -4 + 6 + 1 = 3
    f(3)=−4(3)3+6(3)2+1=−4(27)+6(9)+1=−108+54+1=−53f(3) = -4(3)^3 + 6(3)^2 + 1 = -4(27) + 6(9) + 1 = -108 + 54 + 1 = -53
  4. Compare Values: We compare the values of f(x)f(x) at the critical points and the endpoints to find the absolute minimum.f(−4)=353f(-4) = 353f(0)=1f(0) = 1f(1)=3f(1) = 3f(3)=−53f(3) = -53The absolute minimum value is −53-53, which occurs at x=3x = 3.

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