Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is (a) strictly increasing (b) strictly decreasing
Solution and Explanation
The given function is f(x) = 2x3 − 3x2 − 36x + 7
\(f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x+2)(x-3)\)
\(\therefore f'(x)=0\)⇒ x=-2, 3
The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e.,
\((-\infin, -2),(-2,3)\) and \((3,\infin)\)
In interval \((-∞,-2)\) and \((3,∞)\), \(f'(x)\) is positive while in interval (-2,3), \(f'(x)\) is negative.
Hence, the given function (f) is strictly increasing in intervals \((-∞,-2)\) and \((3,∞)\), while function (f) is strictly decreasing in interval (−2, 3).
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Notes on Applications of Derivatives
Concepts Used:
Increasing and Decreasing Functions
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)
Graphical Representation of Increasing and Decreasing Functions
