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Notely Maths 12
Class 12 Maths
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2.9 Q-2

Question Statement

Find the extreme values of the following functions:

  1. f(x)=1−x3f(x) = 1 - x^3
  2. f(x)=x2−x−2f(x) = x^2 - x - 2
  3. f(x)=5x2−6x+2f(x) = 5x^2 - 6x + 2
  4. f(x)=3x2f(x) = 3x^2
  5. f(x)=3x2−4x+5f(x) = 3x^2 - 4x + 5
  6. f(x)=2x3−2x3−36x+3f(x) = 2x^3 - 2x^3 - 36x + 3
  7. f(x)=x4−4x2f(x) = x^4 - 4x^2

Background and Explanation

To find the extreme values (minima or maxima), we need to:

  1. Compute the first derivative f′(x)f'(x) and set it to zero to find critical points.
  2. Use the second derivative test f′′(x)f''(x) to determine the nature of these critical points:
    • If f′′(x)>0f''(x) > 0, it indicates a relative minimum.
    • If f′′(x)<0f''(x) < 0, it indicates a relative maximum.
    • If f′′(x)=0f''(x) = 0, further investigation is needed.

Solution

1. f(x)=1−x3f(x) = 1 - x^3

  1. Compute f′(x)=−3x2f'(x) = -3x^2.
  2. Set f′(x)=0f'(x) = 0:
−3x2=0  ⟹  x=0 -3x^2 = 0 \implies x = 0
  1. Compute f′′(x)=−6xf''(x) = -6x. At x=0x = 0, f′′(0)=0f''(0) = 0, so the second derivative test is inconclusive.
  2. Observing f′(x)=−3x2f'(x) = -3x^2, it does not change sign at x=0x = 0. Thus, x=0x = 0 is a point of inflection, not an extremum.

2. f(x)=x2−x−2f(x) = x^2 - x - 2

  1. Compute f′(x)=2x−1f'(x) = 2x - 1.
  2. Set f′(x)=0f'(x) = 0:
2x−1=0  ⟹  x=12 2x - 1 = 0 \implies x = \frac{1}{2}
  1. Compute f′′(x)=2f''(x) = 2. Since f′′(x)>0f''(x) > 0, x=12x = \frac{1}{2} is a relative minimum.
  2. Evaluate f(12)f\left(\frac{1}{2}\right):
f(12)=(12)2−12−2=14−12−2=−94 f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 2 = \frac{1}{4} - \frac{1}{2} - 2 = -\frac{9}{4}

Result: Relative minimum at (12,−94)\left(\frac{1}{2}, -\frac{9}{4}\right).

3. f(x)=5x2−6x+2f(x) = 5x^2 - 6x + 2

  1. Compute f′(x)=10x−6f'(x) = 10x - 6.
  2. Set f′(x)=0f'(x) = 0:
10x−6=0  ⟹  x=35 10x - 6 = 0 \implies x = \frac{3}{5}
  1. Compute f′′(x)=10f''(x) = 10. Since f′′(x)>0f''(x) > 0, x=35x = \frac{3}{5} is a relative minimum.
  2. Evaluate f(35)f\left(\frac{3}{5}\right):
f(35)=5(35)2−6(35)+2=95−185+2=15 f\left(\frac{3}{5}\right) = 5\left(\frac{3}{5}\right)^2 - 6\left(\frac{3}{5}\right) + 2 = \frac{9}{5} - \frac{18}{5} + 2 = \frac{1}{5}

Result: Relative minimum at (35,15)\left(\frac{3}{5}, \frac{1}{5}\right).

4. f(x)=3x2f(x) = 3x^2

  1. Compute f′(x)=6xf'(x) = 6x.
  2. Set f′(x)=0f'(x) = 0:
6x=0  ⟹  x=0 6x = 0 \implies x = 0
  1. Compute f′′(x)=6f''(x) = 6. Since f′′(x)>0f''(x) > 0, x=0x = 0 is a relative minimum.
  2. Evaluate f(0)f(0):
f(0)=3(0)2=0 f(0) = 3(0)^2 = 0

Result: Relative minimum at (0,0)(0, 0).

5. f(x)=3x2−4x+5f(x) = 3x^2 - 4x + 5

  1. Compute f′(x)=6x−4f'(x) = 6x - 4.
  2. Set f′(x)=0f'(x) = 0:
6x−4=0  ⟹  x=23 6x - 4 = 0 \implies x = \frac{2}{3}
  1. Compute f′′(x)=6f''(x) = 6. Since f′′(x)>0f''(x) > 0, x=23x = \frac{2}{3} is a relative minimum.
  2. Evaluate f(23)f\left(\frac{2}{3}\right):
f(23)=3(23)2−4(23)+5=43−83+5=113 f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 5 = \frac{4}{3} - \frac{8}{3} + 5 = \frac{11}{3}

Result: Relative minimum at (23,113)\left(\frac{2}{3}, \frac{11}{3}\right).

Follow Same method for part vi and vii

Key Formulas or Methods Used

  1. First derivative test: Solve f′(x)=0f'(x) = 0 to find critical points.
  2. Second derivative test: Use f′′(x)f''(x) to determine the nature of critical points.
  3. Evaluating the function at critical points to find extreme values.

Summary of Steps

  1. Compute the first derivative f′(x)f'(x) and solve f′(x)=0f'(x) = 0.
  2. Use the second derivative f′′(x)f''(x) to classify critical points.
  3. Evaluate f(x)f(x) at critical points to find the extreme values.