Let f(x) = -x^4 - 7 x^3 + 9 x + 5.

Question

Find the open intervals on which f is concave up (down).

Answer 1

The function is f(x) = - x⁴ - 7x³ + 9x + 5.

Differentiate with respect to x.

f'(x) = - 4x³ - 21x² + 9.

Again differentiate with respect to x.

f''(x) = - 12x² - 42x.

Find the x-values at which f''(x) = 0 or f''(x) does not exist.

- 12x² - 42x = 0

- 6x(2x - 7) = 0

- 6x = 0 and 2x - 7 = 0

x = 0 and x = 7/2.

The test intervals are (-∞, 0), (0, 7/2) and (7/2, ∞).

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

(-∞, 0)      x = -1      f''(-1) = - 12(-1)² - 42(-1) = 30 > 0       Concave upward.

(0, 7/2)     x = 1       f''(1) = - 12(1)² - 42(1) = -54 < 0          Concave downward.

(7/2, ∞)    x = 4       f''(4) = - 12(4)² - 42(4) = -360 < 0        Concave downward.

Comments

how did -12x^2-42x become -6x(2x-7)?

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Please, check the updated answer.

Answer 2

The function is f(x) = - x⁴ - 7x³ + 9x + 5.

Differentiate with respect to x.

f'(x) = - 4x³ - 21x² + 9.

Again differentiate with respect to x.

f''(x) = - 12x² - 42x.

Find the x-values at which f''(x) = 0 or f''(x) does not exist.

- 12x² - 42x = 0

- 6x(2x + 7) = 0

- 6x = 0 and 2x + 7 = 0

x = 0 and x = -7/2.

The test intervals are (-∞, -7/2), (-7/2, 0) and (0, ∞).

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

(-∞, -7/2)  x = -4       f''(-4) = - 12(-4)² - 42(-4) = -24 < 0      Concave downward.

(-7/2, 0)     x = -1      f''(-1) = - 12(-1)² - 42(-1) = 30 > 0        Concave upward.

(0, ∞)        x = 1        f''(1) = - 12(1)² - 42(1) = -54 < 0          Concave downward.