Determine if the function is concave up or concave down & identify locations of any inflection points.

Question

Use a sign chart for f" to determine the intervals on which each function f is concave up or concave down, nd identify the locations of any inflection points. Verify your algebraic answers with graphs 

 

1.) f(x) = (x-3)^3 (x-1)

2.) f(x)= square root x / x-2

Answer 1

1) The function f(x) = (x - 3)³(x - 1)

Differentiate with respect to x.

f'(x) = (x - 3)³(1) + (x - 1)3(x - 3)² (1)

f'(x) = (x - 3)³ + (3x - 3)(x² - 6x + 9)

Apply the formula (a - b)³ = a³ - 3a²b + 3ab² - b³

f'(x) = x³ - 9x² + 27x - 27  + 3x³ - 18x² + 27x - 3x² + 18x - 27

f'(x) = 4x³ - 30x² + 72x - 54

Again differentiate with respect to x.

f''(x) = 12x²  - 60x + 72

Equate second derivative = 0

12x²  - 60x + 72 = 0

12(x²  - 5x + 6) = 0

x²  - 5x + 6 = 0

Solve the above equation by using factorization.

x² - 2x - 3x + 6 = 0

x(x - 2) - 3(x - 2) = 0

(x - 2)(x - 3) = 0

x = 2 and x = 3

The test intervals are (-∞, 2), (2, 3) and (3, ∞).

Interval  Test Value                      Sign of f''(x)                           Conclusion

(-∞, 2)   x = -1      f''(-1) = 12(-1)² - 60(-1) + 72 = 144 > 0      Concave upward.

(2, 3)     x = 2.5    f''(2.5) = 12(2.5)² - 60(2.5) + 72 = -3 < 0  Concave downward.

(3, ∞)   x = 4       f''(4) = 12(4)² - 60(4) + 72 = 24 > 0          Concave upward.

 

To find inflection points, using the x - values find the corresponding y - value with the curve.

Now Substitute the values x = 2 and 3 in original function.

f(x) = (2 - 3)³(2 - 1)

y = - 1(1)

y = - 1

f(x) = (3 - 3)³(2 - 1)

y = 0

Inflection points are (2, - 1) and (3, 0)

Graph