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Magazine Chapter 5: Problem 14
If \(f(x)=\left\\{\begin{array}{cc}3 x^{2}+12 x-1 ; & -1 \leq x \leq 2 \\ 37-x
; & 2
Short Answer
Expert verified
The short answer is: (a) True, (b) True, (c) True, (d) False.
Step by step solution
01
Analyze increasing behavior in [-1,2]
For the interval [-1,2], the function is defined as \(f(x) = 3x^2 + 12x - 1\). To see if the function is increasing, we need to take its derivative and check if it's positive.
\(\frac{df(x)}{dx} = 6x + 12\)
Now, let's check its sign in the interval:
For \(x \in [-1,2]\), \(6x + 12\) is always positive as the minimum value occurs at x=-1: \(6(-1) + 12 = 6 > 0\)
Since the derivative is always positive in [-1,2], \(f(x)\) is increasing in this interval. So, option (a) is true.
02
Check continuity in [-1,3]
First, we need to ensure that the function is continuous at the point x=2. For left continuity, we want \( \lim_{x\to 2^{-}} f(x) = f(2) \) and for right continuity, we want \( \lim_{x\to 2^{+}} f(x) = f(2) \).
For the left limit, using the quadratic part of the given function:
\(\lim_{x\to 2^{-}} f(x) = 3(2)^2 + 12(2) - 1 = 35\)
For the right limit, using the linear part of the given function:
\(\lim_{x\to 2^{+}} f(x) = 37 - 2 = 35\)
Since both left and right limits are equal to 35, the function is continuous at x = 2.
Both parts of the function are continuous in their respective intervals, so the function is continuous on [-1,3]. Therefore, option (b) is true.
03
Check the existence of f'(2) (differentiability)
We already have the derivative for the quadratic part \(\frac{df(x)}{dx} = 6x + 12\).
For the linear part, its derivative is constant: \(\frac{df(x)}{dx} = -1\).
Now, we need to check if the left and right derivatives are equal at x=2:
Left derivative: \(6(2) + 12 = 24\)
Right derivative: \(-1\)
Since the left and right derivatives are not equal at x=2, the function is not differentiable at x=2. Thus, option (c) is true.
04
Check if f(x) has maximum value at x=2
We know that the function is increasing in the interval [-1,2] from our analysis in Step 1. Since the function is not differentiable at x=2, it has neither maximum nor minimum at x=2. Therefore, option (d) is false.
In conclusion, options (a), (b), and (c) are true, and option (d) is false.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity
When we talk about continuity, it means that a function has no breaks, jumps, or holes at any point in its domain. For a function to be continuous at a point, the left-hand limit and right-hand limit as we approach the point must both equal the function's value at that point.
In our exercise, function \(f(x)\) transitions from a quadratic form \(3x^2 + 12x - 1\) on \([-1, 2]\) to a linear form \(37 - x\) on \((2, 3]\). To confirm continuity at \(x = 2\), we calculate the left and right limits and ensure they equal \(f(2)\).
In our exercise, function \(f(x)\) transitions from a quadratic form \(3x^2 + 12x - 1\) on \([-1, 2]\) to a linear form \(37 - x\) on \((2, 3]\). To confirm continuity at \(x = 2\), we calculate the left and right limits and ensure they equal \(f(2)\).
- The left limit leading up to \(x = 2\) from the quadratic part results in \(35\).
- The right limit going away from \(x = 2\) using the linear part also gives \(35\).
- Because both limits equal \(f(2) = 35\), \(f(x)\) is continuous at \(x = 2\).
Differentiability
Differentiability refers to the ability of a function to have a derivative at a certain point. For a function to be differentiable at a point, its left-hand derivative must equal its right-hand derivative as we approach that point.
In our exercise, we checked differentiability at \(x = 2\), where the function changes from a quadratic to a linear form. We need to see if the slope coming from different segments of \(f(x)\) are the same at this transition:
In our exercise, we checked differentiability at \(x = 2\), where the function changes from a quadratic to a linear form. We need to see if the slope coming from different segments of \(f(x)\) are the same at this transition:
- The left-hand derivative, from the quadratic side, is \(6x + 12\), which computes to \(24\) at \(x = 2\).
- The right-hand derivative, from the linear side, is \(-1\), a constant, not changing with \(x\).
- Since the left and right derivatives are not equal at \(x = 2\), the function is not differentiable there.
Increasing Function
An increasing function is one where, as you move from left to right along the x-axis, the function values rise. This is indicated by its derivative being positive in a given interval.
For the interval \([-1, 2]\), the function takes the form \(3x^2 + 12x - 1\). The derivative of this is \(6x + 12\).
For the interval \([-1, 2]\), the function takes the form \(3x^2 + 12x - 1\). The derivative of this is \(6x + 12\).
- We find \( rac{df(x)}{dx} = 6x + 12\) which results in a positive derivative within \([-1,2]\), starting with a minimum value of \(6\) at \(x = -1\).
- Since the derivative \(6x + 12\) remains positive throughout \([-1, 2]\), \(f(x)\) is conclusively increasing over this range.
Maximum Value
The maximum value of a function in an interval is the highest point the function reaches within that interval. To determine whether the function \(f(x)\) reaches its maximum at \(x = 2\), we explore its behavior on \([-1, 3]\).
From earlier, we know:
From earlier, we know:
- \(f(x)\) is increasing up to \(x = 2\).
- We also noted that the function transitions at \(x = 2\) to a section where it''s decreasing due to a constant derivative of \(-1\) for \((2, 3]\).
- However, though increasing up to \(x=2\), differentiability issues arise at this point, and analyzing limit behavior indicates the value doesn鈥檛 necessarily peak here.
