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).
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is
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)