If a function $ f(x) = x^2 + bx + 1 $ is increasing in the interval $[1, 2]$, then the least value of $ b $ is:

Question

Prepp Team · Accepted Answer

Finding the Least Value of b for an Increasing Quadratic Function

We are given a function $ f(x) = x^2 + bx + 1 $. We are told that this function is increasing in the interval $ [1, 2] $. We need to find the least possible value of the constant $ b $.

Condition for an Increasing Function

A function $ f(x) $ is increasing over a given interval if its derivative, $ f'(x) $, is greater than or equal to zero ($ f'(x) \ge 0 $) for all $ x $ in that interval.

Calculate the Derivative

First, let's find the derivative of the given function $ f(x) $:

$ f(x) = x^2 + bx + 1 $

Using the power rule for differentiation, $ \frac{d}{dx}(x^n) = nx^{n-1} $, and the rule for a constant multiple $ \frac{d}{dx}(cx) = c $, and the rule for a constant $ \frac{d}{dx}(c) = 0 $:

$ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(bx) + \frac{d}{dx}(1) $

$ f'(x) = 2x^{2-1} + b \cdot 1 + 0 $

$ f'(x) = 2x + b $

Applying the Increasing Function Condition in the Interval

For the function $ f(x) $ to be increasing in the interval $ [1, 2] $, its derivative $ f'(x) $ must be non-negative for all $ x $ in this interval. So, we must have:

$ f'(x) \ge 0 $ for all $ x \in [1, 2] $

$ 2x + b \ge 0 $ for all $ x \in [1, 2] $

Determining the Range for b

We need to find the values of $ b $ that satisfy the inequality $ 2x + b \ge 0 $ for all $ x $ in the interval $ [1, 2] $. We can rearrange the inequality to isolate $ b $:

$ b \ge -2x $

This inequality must hold for every value of $ x $ in the interval $ [1, 2] $. This means that $ b $ must be greater than or equal to the maximum possible value of $ -2x $ within this interval.

Let's consider the function $ g(x) = -2x $. This is a linear function with a negative slope ($-2$), which means it is a decreasing function. A decreasing function attains its maximum value at the smallest point in the interval.

The interval is $ [1, 2] $. The smallest value in this interval is $ x = 1 $.

The maximum value of $ -2x $ in the interval $ [1, 2] $ occurs at $ x = 1 $:

Maximum value $ = -2 \times 1 = -2 $

So, the condition $ b \ge -2x $ for all $ x \in [1, 2] $ is equivalent to requiring that $ b $ must be greater than or equal to the maximum value of $ -2x $ in $ [1, 2] $. Therefore, we must have:

$ b \ge -2 $

Finding the Least Value of b

The inequality $ b \ge -2 $ means that $ b $ can be $ -2 $ or any value greater than $ -2 $. The set of possible values for $ b $ is $ [-2, \infty) $.

We are looking for the least value of $ b $ that satisfies this condition. The smallest value in the set $ [-2, \infty) $ is $ -2 $.

Therefore, the least value of $ b $ is $ -2 $.

Revision Table: Key Concepts

Concept	Explanation	Relevance to Problem
Increasing Function	A function $ f(x) $ is increasing on an interval if for any $ x_1 < x_2 $ in the interval, $ f(x_1) \le f(x_2) $.	Given condition for $ f(x) $ in $ [1, 2] $.
Derivative $ f'(x) $	Measures the instantaneous rate of change of $ f(x) $. For a smooth function, $ f(x) $ is increasing where $ f'(x) \ge 0 $.	Used to determine the condition for the function increasing.
Quadratic Function	A function of the form $ ax^2 + bx + c $. The given function is a quadratic with $ a=1, c=1 $.	The specific type of function being analyzed.
Interval $ [1, 2] $	The specific range of $ x $ values where the function is increasing.	Constraints the application of the $ f'(x) \ge 0 $ condition.

Additional Information: General Case of Quadratic Functions

Consider a general quadratic function $ f(x) = ax^2 + bx + c $. The derivative is $ f'(x) = 2ax + b $.

The vertex of the parabola is at $ x = -\frac{b}{2a} $.

If $ a > 0 $ (parabola opens upwards), the function is decreasing for $ x < -\frac{b}{2a} $ and increasing for $ x > -\frac{b}{2a} $. The function is increasing on the interval $ [-\frac{b}{2a}, \infty) $.
If $ a < 0 $ (parabola opens downwards), the function is increasing for $ x < -\frac{b}{2a} $ and decreasing for $ x > -\frac{b}{2a} $. The function is increasing on the interval $ (-\infty, -\frac{b}{2a}] $.

In our specific problem, $ f(x) = x^2 + bx + 1 $, so $ a = 1 $ (which is > 0). The function is increasing for $ x \ge -\frac{b}{2(1)} $, i.e., $ x \ge -\frac{b}{2} $.

We are given that the function is increasing in the interval $ [1, 2] $. This means the interval $ [1, 2] $ must be a sub-interval of the increasing interval $ [-\frac{b}{2}, \infty) $.

For $ [1, 2] $ to be a sub-interval of $ [-\frac{b}{2}, \infty) $, the starting point of $ [1, 2] $ must be greater than or equal to the starting point of $ [-\frac{b}{2}, \infty) $. That is,

$ 1 \ge -\frac{b}{2} $

Multiply both sides by 2:

$ 2 \ge -b $

Multiply both sides by -1 and reverse the inequality sign:

$ -2 \le b $

or $ b \ge -2 $

This confirms our earlier result obtained using the derivative condition $ f'(x) \ge 0 $. The least value of $ b $ is indeed $ -2 $.

Prepp Team · Answer

5

Prepp Team · Answer

0

Prepp Team · Answer

-4

Concept	Explanation	Relevance to Problem
Increasing Function	A function \( f(x) \) is increasing on an interval if for any \( x_1 < x_2 \) in the interval, \( f(x_1) \le f(x_2) \).	Given condition for \( f(x) \) in \( [1, 2] \).
Derivative \( f'(x) \)	Measures the instantaneous rate of change of \( f(x) \). For a smooth function, \( f(x) \) is increasing where \( f'(x) \ge 0 \).	Used to determine the condition for the function increasing.
Quadratic Function	A function of the form \( ax^2 + bx + c \). The given function is a quadratic with \( a=1, c=1 \).	The specific type of function being analyzed.
Interval \( [1, 2] \)	The specific range of \( x \) values where the function is increasing.	Constraints the application of the \( f'(x) \ge 0 \) condition.

List-I	List-II
(A) Absolute maximum value	(I) 3
(B) Absolute minimum value	(II) 0
(C) Point of maxima	(III) -5
(D) Point of minima	(IV) 4