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Question: Let $f(x)$ be a cubic polynomial with $f(1) = -10, f(-1) = 6$, and has a local minima at $x = 1$, an...

Let f(x)f(x) be a cubic polynomial with f(1)=−10,f(−1)=6f(1) = -10, f(-1) = 6, and has a local minima at x=1x = 1, and f′(x)f'(x) has a local minima at x=−1x = -1 then ∣f(2)∣|f(2)| is equal to ____

Answer

3

Explanation

Solution

Let the cubic polynomial be f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d.

First, we find the derivatives of f(x)f(x): f′(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c f′′(x)=6ax+2bf''(x) = 6ax + 2b f′′′(x)=6af'''(x) = 6a

We are given the following conditions:

  1. f(x)f(x) has a local minima at x=1x = 1: This implies f′(1)=0f'(1) = 0 and f′′(1)>0f''(1) > 0. f′(1)=3a(1)2+2b(1)+c=3a+2b+c=0f'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c = 0 (Equation 1)

  2. f′(x)f'(x) has a local minima at x=−1x = -1: For f′(x)f'(x) to have a local minimum, its first derivative, f′′(x)f''(x), must be zero at x=−1x = -1, and its second derivative, f′′′(x)f'''(x), must be positive at x=−1x = -1. f′′(−1)=6a(−1)+2b=−6a+2b=0f''(-1) = 6a(-1) + 2b = -6a + 2b = 0 From this, 2b=6a  ⟹  b=3a2b = 6a \implies b = 3a. (Equation 2) f′′′(−1)=6af'''(-1) = 6a. For a local minimum, f′′′(−1)>0f'''(-1) > 0, so 6a>0  ⟹  a>06a > 0 \implies a > 0.

Now, substitute b=3ab = 3a into Equation 1: 3a+2(3a)+c=03a + 2(3a) + c = 0 3a+6a+c=03a + 6a + c = 0 9a+c=0  ⟹  c=−9a9a + c = 0 \implies c = -9a. (Equation 3)

Now we have bb and cc in terms of aa. Let's substitute these into the polynomial f(x)f(x): f(x)=ax3+(3a)x2+(−9a)x+df(x) = ax^3 + (3a)x^2 + (-9a)x + d f(x)=ax3+3ax2−9ax+df(x) = ax^3 + 3ax^2 - 9ax + d

We are given two more conditions:

  1. f(1)=−10f(1) = -10: Substitute x=1x=1 into f(x)f(x): a(1)3+3a(1)2−9a(1)+d=−10a(1)^3 + 3a(1)^2 - 9a(1) + d = -10 a+3a−9a+d=−10a + 3a - 9a + d = -10 −5a+d=−10  ⟹  d=5a−10-5a + d = -10 \implies d = 5a - 10. (Equation 4)

  2. f(−1)=6f(-1) = 6: Substitute x=−1x=-1 into f(x)f(x): a(−1)3+3a(−1)2−9a(−1)+d=6a(-1)^3 + 3a(-1)^2 - 9a(-1) + d = 6 −a+3a+9a+d=6-a + 3a + 9a + d = 6 11a+d=611a + d = 6. (Equation 5)

Now we have a system of two linear equations for aa and dd. Substitute Equation 4 into Equation 5: 11a+(5a−10)=611a + (5a - 10) = 6 16a−10=616a - 10 = 6 16a=1616a = 16 a=1a = 1

Now, find the values of b,c,db, c, d using a=1a=1: b=3a=3(1)=3b = 3a = 3(1) = 3 c=−9a=−9(1)=−9c = -9a = -9(1) = -9 d=5a−10=5(1)−10=5−10=−5d = 5a - 10 = 5(1) - 10 = 5 - 10 = -5

So, the cubic polynomial is f(x)=x3+3x2−9x−5f(x) = x^3 + 3x^2 - 9x - 5.

Finally, we need to find ∣f(2)∣|f(2)|: f(2)=(2)3+3(2)2−9(2)−5f(2) = (2)^3 + 3(2)^2 - 9(2) - 5 f(2)=8+3(4)−18−5f(2) = 8 + 3(4) - 18 - 5 f(2)=8+12−18−5f(2) = 8 + 12 - 18 - 5 f(2)=20−23f(2) = 20 - 23 f(2)=−3f(2) = -3

Therefore, ∣f(2)∣=∣−3∣=3|f(2)| = |-3| = 3.

The condition f′′(1)>0f''(1) > 0 for local minima at x=1x=1 is also satisfied: f′′(x)=6ax+2bf''(x) = 6ax + 2b. With a=1,b=3a=1, b=3: f′′(x)=6x+6f''(x) = 6x+6. f′′(1)=6(1)+6=12f''(1) = 6(1)+6 = 12, which is >0>0.