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# Consider the equation below. f(x) = 2x^{3} + 3x^{2} - 72x

(a) Find the interval on which f is increasing. Find the interval on which f is decreasing.

(b) Find the local minimum and maximum values of f.

(c) Find the inflection point. (x, y) = find the interval on which f is concave up. Find the interval on which f is concave down.

**Solution:**

Given function f(x) = 2x³ + 3x² - 72x

Diff .w.r. to x

fꞌ(x) = 6x² + 6x - 72x

= 6( x² + x -12)

= 6(x + 4) (x - 3)

Now consider fꞌ(x) = 0,

⇒ 6(x + 4) (x - 3) = 0

⇒ x = 3 or x = -4

Now, divide the entire real set into disjoint intervals as (-∞, -4) , (-4, 3) and (3, ∞)

(i) When x ∊ (-∞, -4) , fꞌ(x) = 6(x + 4) (x - 3)

= (+Ve)(-ve)(-ve) = +ve

Since fꞌ(x) is positive f(x) is increasing in (-∞, -4)

(ii) When x ∊ (-4, 3) fꞌ(x) = 6(x + 4) (x - 3)

= (+Ve)(+ve)(-ve) = -ve

Since fꞌ(x) is negative f(x) is decreasing in ( -4, 3)

(iii) When x ∊ (3, ∞) , fꞌ(x) = 6(x + 4) (x - 3)

= (+Ve)(+ve)(+ve) = +ve

Since fꞌ(x) is positive f(x) is increasing in(3, ∞)

∴ f(x) is increasing in (-∞, -4) ∪ (3, ∞)

And f(x) is decreasing in (-4, 3)

(b) Given f(x) = 2x³ + 3x² - 72x

Diff .w.r. to x

fꞌ(x) = 6x² + 6x - 72x

= 6( x² + x -12)

= 6(x + 4) (x - 3)

Now consider fꞌ(x) = 0,

⇒ 6(x + 4) (x - 3) = 0

⇒ x = 3 or x = -4

Now differentiate again w.r.to x

fꞌꞌ(x) = 12 x + 6

When x = -4, fꞌꞌ(-4) = 12(-4) + 6 = - 42 < 0

∴ f(x) is maximum when x = -4 and max value is f(-4) = 2(-4)³ + 3(-4)² -72(-4) = 208

When x = 3 , fꞌꞌ(3) = 12(3) + 6 = 42 > 0

∴f(x) is minimum when x = 3 and minimum value is f(3) = 2(3)³ + 3(3)² -72(3) =-135

(c) For point of inflection condition is fꞌꞌ(x) = 0 , and fꞌꞌꞌ(x)≠0

Consider fꞌꞌ(x) = 0 ⇒ 12x + 6 = 0 ⇒ x = -1/2 also fꞌꞌꞌ(x) = 12

And fꞌꞌꞌ(-1/2) = 12 ≠0

And when x = -1/2 .

Y = f(-1/2) = 2(-1/2)³ + 3(-1/2)² - 72(-1/2) = 73/2

∴ Point of inflection is (-1/2, 73/2)

## Consider the equation below. f(x) = 2x^{3} + 3x^{2} - 72x. (a) Find the interval on which f is increasing. Find the interval on which f is decreasing. (b) Find the local minimum and maximum values of f. (c) Find the inflection point. (x, y) = find the interval on which f is concave up. Find the interval on which f is concave down.

**Summary:**

In the given equation f(x) = 2x³ + 3x² - 72x,f(x) is increasing in (-∞, -4) ∪ (3, ∞) and f(x) is decreasing in (-4, 3), f(x) is maximum when x = -4, f(x) is minimum when x = 3 and Point of inflection is (-1/2, 73/2). And f(x) is concave up in the interval (-4, 3)

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