How do I solve this calculus problem?

Question

Let f(x)= 2 x^4-4 x^2+6 

Input the interval(s) on which f is increasing.: 

Input the interval(s) on which f is decreasing.: 

Find the point(s) at which f achieves a local maximum.: 

Find the point(s) at which f achieves a local minimum.: 

Find the interval(s) on which f is concave up.: 

Find the interval(s) on which f is concave down.:

Find all inflection points.: 

Can someone please help?

Answer 1

The function is f(x) = 2x⁴ - 4x² + 6.

f^,(x) = 8x³ - 8x.

f^,,(x) = 24x² - 8.

TEST FOR INCREASING AND DECREASING FUNCTIONS :

If f^,(x) > 0 (Positive) for all x in (a, b), then f(x) is increasing on [a, b].

If f^,(x) < 0 (Negative) for all x in (a, b), then f(x) is decreasing on [a, b].

To find the critical or key numbers, to make the first derivative equal to zero or f ' (x) = 0.

f^,(x) = 8x³ - 8x = 8x(x² - 1) = 0.

The key numbers are x = 0 and x = ± 1.

Test intervals    x - Value                Polynomial or f^,(x) value                   Conclusion

(- ∞, - 1)            x = - 2            8(- 2)³ - 8(- 2) = - 64 + 16 = - 48 < 0        Decreasing

(- 1, 0)               x = - 0.5         8(- 0.5)³ - 8(- 0.5) = - 1 + 4 = 3 > 0         Increasing

(0, 1)                 x = 0.5           8(0.5)³ - 8(0.5) = 1 - 4 = - 3 < 0              Decreasing

(1, ∞)                x = 2              8(2)³ - 8(2) = 64 - 16 = 48 > 0                Increasing

So, f(x) is increasing on the interval (- 1, 0) and (1, ∞) and decreasing on the interval (- ∞, - 1) and (0, 1).

 

Find Extrema :

To find out extrema, use theorem.

If f " (x) > 0 (positive) ------> minimum point.

If f " (x) < 0 (negative) ------> maximum point.

f^,(x) = 8x³ - 8x = 8x(x² - 1) = 0.

The key numbers are x = 0 and x = ± 1.

So, lets plug each critical point in f " (x) = 24x² - 8.

If x = - 1 then f " (- 1) = 24(- 1)² - 8 = 24 - 8 = 16 > 0 (positive), therefore local minimum point.

If x = 0 then f " (0) = 24(0)² - 8 = - 8 < 0 (negative), therefore local maximum point.

If x = 1 then f " (1) = 24(1)² - 8 = 24 - 8 = 16 > 0 (positive), therefore local minimum point.

To find the f(x) to each x for local max and local min plugging those values in the original function.

If x = - 1 then, f(- 1) = 2(- 1)⁴ - 4(- 1)² + 6 = 2 - 4 + 6 = 4.

If x = 0 then, f(0) = 2(0)⁴ - 4(0)² + 6 = 6.

If x = 1 then, f(1) = 2(1)⁴ - 4(1)² + 6 = 2 - 4 + 6 = 4.

The relative maximum point is (0, 6) and the relative minimum points are (- 1, 4) and (1, 4).

 

TEST FOR CONCAVITY :

If f " (x) > 0 (Positive) for all x in (a, b), then f(x) is concave upward on (a, b).

If f " (x) < 0 (Negative) for all x in (a, b), then f(x) is concave downward on (a, b).

To find the Concavity, locate the x - values at which f " (x) = 0 or f " (x) does not exist.

f " (x) = 24x² - 8 = 0 ⟹ 24x² = 8 ⟹ x² = 1/3 ⟹ x = ± 1/√3.

Test intervals         x - Value                Polynomial or f " (x) value                   Conclusion

(- ∞, - 1/√3)            x = - 1            24(- 1)² - 8 = 24 - 8 = 16 > 0                Concave upward

(- 1/√3, 1/√3)          x = 0                   24(0)² - 8 = - 8 < 0                         Concave downward

(1/√3, ∞)                x = 1              24(1)² - 8 = 24 - 8 = 16 > 0                  Concave upward.

 

Points of Inflection : If (c, f(c)) is a point of inflection of the graph of f(x), then either f " (c) = 0 or f " (x) does not exist.

f " (x) = 24x² - 8 = 0 ⟹ 24x² = 8 ⟹ x² = 1/3 ⟹ x = ± 1/√3.

If x = - 1/√3 then, f(- 1/√3) = 2(- 1/√3)⁴ - 4(- 1/√3)² + 6 = 2/9 - 4/3 + 6 = (2 - 12 + 54)/9 = 44/9.

If x = 1/√3 then, f(1/√3) = 2(1/√3)⁴ - 4(1/√3)² + 6 = 2/9 - 4/3 + 6 = (2 - 12 + 54)/9 = 44/9.

The Inflection points are (- 1/√3, 44/9) and (1/√3, 44/9).