The interval in which the function $f(x) = x^3 - 6x^2 + 9x + 10$ is increasing in
- [1, 3]
- $( - \infty , 1) \cup (3, \infty)$
- $( - \infty , - 1] \cup [3, \infty)$
- $( - \infty , 1] \cup [3, \infty)$
The Correct Option is D
Solution and Explanation
$f '\left(x\right)=3x^{2}-12x+9$
$f '\left(x\right)=0$
$3x^{2}-12x+9=0 \quad$(dividing by $3$)
$x^{2}-4x+3=0$
$x^{2}-3x-x+3=0$
$x\left(x-3\right)-1\left(x-3\right)=0$
$x=3, 1$
$(-\infty, 1 ] \cup[3,\infty)$
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Questions Asked in KCET exam
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \).
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- The function f(x) is given by:
For x < 0:
f(x) = ex + axFor x ≥ 0:
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- The function \( f(x) = \tan x - x \)
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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
