Find the least value of a such that the function f given \(f(x)=x^2+ax+1\) is strictly increasing on \((1, 2)\).
Find the least value of a such that the function f given \(f(x)=x^2+ax+1\) is strictly increasing on \((1, 2)\).
Solution and Explanation
We have,
f'(x) = x2+ax+1
f'(x) = 2x+a
Now, function f will be increasing in (1, 2), if f'(x)>0 in (1, 2).
f'(x)>0
⇒ 2x + a > 0
⇒ 2x > −a
⇒ x>-\(\frac a2\)
Therefore, we have to find the least value of a such that
x>-\(\frac a2\), when x ε (1,2).
x>-\(\frac a2\), (when 1<x<2)
Thus, the least value of a for f to be increasing on (1, 2) is given by,
-\(\frac a2\) = 1
⇒ a=-2
Hence, the required value of a is −2.
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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
