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Given f'(x) = x2 + 2x − 3, find the turning points and intervals of increase/decrease of f(x)?

Mathematics

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Answer

a

(−3,9+C) and (1,−53+C)(-3, 9+C) \text{ and } (1, -\frac{5}{3}+C)

Step-by-step solution

5 steps

The given function is the derivative of f(x)f(x):

f′(x)=x2+2x−3f'(x) = x^2 + 2x - 3

Part a· 3 steps

Determine coordinates of the turning point.

  1. Find the xx-coordinates of the turning points of f(x)f(x) by setting f′(x)=0f'(x) = 0.

    x2+2x−3=0x^2 + 2x - 3 = 0

    Factor the quadratic equation:

    (x+3)(x−1)=0(x+3)(x-1) = 0

    This gives two xx-coordinates for the turning points:

    x=−3orx=1x = -3 \quad \text{or} \quad x = 1
  2. Find the function f(x)f(x) by integrating f′(x)f'(x).

    f(x)=∫(x2+2x−3) dxf(x) = \int (x^2 + 2x - 3)\,dx f(x)=x2+12+1+2x1+11+1−3x+Cf(x) = \frac{x^{2+1}}{2+1} + \frac{2x^{1+1}}{1+1} - 3x + C f(x)=x33+x2−3x+Cf(x) = \frac{x^3}{3} + x^2 - 3x + C
  3. Substitute the xx-coordinates into f(x)f(x) to find the corresponding yy-coordinates.

    For x=−3x = -3:

    f(−3)=(−3)33+(−3)2−3(−3)+Cf(-3) = \frac{(-3)^3}{3} + (-3)^2 - 3(-3) + C f(−3)=−273+9+9+Cf(-3) = \frac{-27}{3} + 9 + 9 + C f(−3)=−9+9+9+Cf(-3) = -9 + 9 + 9 + C f(−3)=9+Cf(-3) = 9 + C

    The first turning point is (−3,9+C)(-3, 9+C).

    For x=1x = 1:

    f(1)=(1)33+(1)2−3(1)+Cf(1) = \frac{(1)^3}{3} + (1)^2 - 3(1) + C f(1)=13+1−3+Cf(1) = \frac{1}{3} + 1 - 3 + C f(1)=13−2+Cf(1) = \frac{1}{3} - 2 + C f(1)=−53+Cf(1) = -\frac{5}{3} + C

    The second turning point is (1,−53+C)(1, -\frac{5}{3}+C).

    The coordinates of the turning points are (−3,9+C) and (1,−53+C)\boxed{(-3, 9+C) \text{ and } (1, -\frac{5}{3}+C)}.

Part b· 2 steps

Determine the interval over which ff is increasing and decreasing.

  1. Recall that f(x)f(x) is increasing when f′(x)>0f'(x) > 0 and decreasing when f′(x)<0f'(x) < 0.

    We found the roots of f′(x)=0f'(x) = 0 to be x=−3x = -3 and x=1x = 1.
    Since f′(x)=x2+2x−3f'(x) = x^2 + 2x - 3 is a parabola opening upwards (the coefficient of x2x^2 is positive), f′(x)f'(x) is positive outside its roots and negative between its roots.

  2. Determine the intervals.

    f(x)f(x) is increasing when f′(x)>0f'(x) > 0:

    x2+2x−3>0x^2 + 2x - 3 > 0 (x+3)(x−1)>0(x+3)(x-1) > 0

    This occurs when x<−3x < -3 or x>1x > 1.

    $f(x

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