Solveeit Logo
Login

Question

Question: Find local maxima and local minima f(x)=e^x(1-x)^3...

Find local maxima and local minima f(x)=e^x(1-x)^3

Answer

Local maximum at x=-2, value 27/e^2. No local minimum.

Explanation

Solution

To find the local maxima and local minima of the function f(x)=ex(1−x)3f(x)=e^x(1-x)^3, we follow these steps:

  1. Find the first derivative, f′(x)f'(x):
    Using the product rule (uv)′=u′v+uv′(uv)' = u'v + uv', where u=exu = e^x and v=(1−x)3v = (1-x)^3:
    u′=exu' = e^x
    v′=3(1−x)2⋅(−1)=−3(1−x)2v' = 3(1-x)^2 \cdot (-1) = -3(1-x)^2
    So,
    f′(x)=ex(1−x)3+ex(−3(1−x)2)f'(x) = e^x(1-x)^3 + e^x(-3(1-x)^2)
    Factor out ex(1−x)2e^x(1-x)^2:
    f′(x)=ex(1−x)2[(1−x)−3]f'(x) = e^x(1-x)^2 [(1-x) - 3]
    f′(x)=ex(1−x)2(−x−2)f'(x) = e^x(1-x)^2 (-x-2)
    f′(x)=−ex(1−x)2(x+2)f'(x) = -e^x(1-x)^2 (x+2)

  2. Find the critical points by setting f′(x)=0f'(x) = 0:
    −ex(1−x)2(x+2)=0-e^x(1-x)^2 (x+2) = 0
    Since ex>0e^x > 0 for all real xx, we must have:
    (1−x)2(x+2)=0(1-x)^2 (x+2) = 0
    This gives two critical points:
    1−x=0  ⟹  x=11-x = 0 \implies x = 1
    x+2=0  ⟹  x=−2x+2 = 0 \implies x = -2

  3. Find the second derivative, f′′(x)f''(x):
    First, expand f′(x)f'(x):
    f′(x)=−ex(x2−2x+1)(x+2)f'(x) = -e^x(x^2 - 2x + 1)(x+2)
    f′(x)=−ex(x3+2x2−2x2−4x+x+2)f'(x) = -e^x(x^3 + 2x^2 - 2x^2 - 4x + x + 2)
    f′(x)=−ex(x3−3x+2)f'(x) = -e^x(x^3 - 3x + 2)
    Now, differentiate f′(x)f'(x) using the product rule again:
    f′′(x)=−[ex(x3−3x+2)+ex(3x2−3)]f''(x) = -[e^x(x^3 - 3x + 2) + e^x(3x^2 - 3)]
    f′′(x)=−ex(x3+3x2−3x+2−3)f''(x) = -e^x(x^3 + 3x^2 - 3x + 2 - 3)
    f′′(x)=−ex(x3+3x2−3x−1)f''(x) = -e^x(x^3 + 3x^2 - 3x - 1)

  4. Apply the second derivative test:

    • At x=−2x = -2:
      Substitute x=−2x=-2 into f′′(x)f''(x):
      f′′(−2)=−e−2((−2)3+3(−2)2−3(−2)−1)f''(-2) = -e^{-2}((-2)^3 + 3(-2)^2 - 3(-2) - 1)
      f′′(−2)=−e−2(−8+3(4)+6−1)f''(-2) = -e^{-2}(-8 + 3(4) + 6 - 1)
      f′′(−2)=−e−2(−8+12+6−1)f''(-2) = -e^{-2}(-8 + 12 + 6 - 1)
      f′′(−2)=−e−2(9)f''(-2) = -e^{-2}(9)
      f′′(−2)=−9e−2f''(-2) = -9e^{-2}
      Since f′′(−2)<0f''(-2) < 0, x=−2x=-2 is a point of local maximum.

    • At x=1x = 1:
      Substitute x=1x=1 into f′′(x)f''(x):
      f′′(1)=−e1(13+3(1)2−3(1)−1)f''(1) = -e^1(1^3 + 3(1)^2 - 3(1) - 1)
      f′′(1)=−e(1+3−3−1)f''(1) = -e(1 + 3 - 3 - 1)
      f′′(1)=−e(0)f''(1) = -e(0)
      f′′(1)=0f''(1) = 0
      Since f′′(1)=0f''(1) = 0, the second derivative test is inconclusive. We use the first derivative test.

  5. Apply the first derivative test for x=1x=1:
    Recall f′(x)=−ex(1−x)2(x+2)f'(x) = -e^x(1-x)^2 (x+2).
    The term (1−x)2(1-x)^2 is always non-negative. exe^x is always positive. So the sign of f′(x)f'(x) depends on −(x+2)-(x+2).

    • For x<1x < 1 (e.g., x=0.5x=0.5): x+2=2.5>0x+2 = 2.5 > 0. So f′(x)=−e0.5(1−0.5)2(2.5)<0f'(x) = -e^{0.5}(1-0.5)^2(2.5) < 0.
    • For x>1x > 1 (e.g., x=1.5x=1.5): x+2=3.5>0x+2 = 3.5 > 0. So f′(x)=−e1.5(1−1.5)2(3.5)<0f'(x) = -e^{1.5}(1-1.5)^2(3.5) < 0.

    Since f′(x)f'(x) does not change sign around x=1x=1 (it remains negative), x=1x=1 is an inflection point, not a local extremum. Therefore, there is no local minimum.

  6. Calculate the local maximum value:
    The local maximum occurs at x=−2x=-2. Substitute this value back into the original function f(x)=ex(1−x)3f(x)=e^x(1-x)^3:
    f(−2)=e−2(1−(−2))3f(-2) = e^{-2}(1-(-2))^3
    f(−2)=e−2(1+2)3f(-2) = e^{-2}(1+2)^3
    f(−2)=e−2(3)3f(-2) = e^{-2}(3)^3
    f(−2)=27e−2f(-2) = 27e^{-2}
    f(−2)=27e2f(-2) = \frac{27}{e^2}

The function has a local maximum at x=−2x=-2 with value 27e2\frac{27}{e^2}. There is no local minimum.