Asked by Fiona
For a real number t, let M(t) denote the maximum value of f(x) = 4x^2 - x + t for 11 \le x \le 1. What is the smallest possible value of M(t)?
Answers
Answered by
GPT-4o mini
To find the maximum value \( M(t) \) of the function \( f(x) = 4x^2 - x + t \) for \( x \) in the interval \( [1, 11] \), we start by determining the critical points of \( f(x) \).
First, we calculate the derivative of \( f(x) \):
\[
f'(x) = \frac{d}{dx}(4x^2 - x + t) = 8x - 1
\]
Next, we set the derivative equal to zero to find the critical points:
\[
8x - 1 = 0 \implies x = \frac{1}{8}
\]
However, since \( \frac{1}{8} \) is not within the interval \( [1, 11] \), we only need to evaluate \( f(x) \) at the endpoints of the interval.
Now, we compute the values of \( f(x) \) at \( x = 1 \) and \( x = 11 \):
1. **Calculate \( f(1) \)**:
\[
f(1) = 4(1)^2 - 1 + t = 4 - 1 + t = 3 + t
\]
2. **Calculate \( f(11) \)**:
\[
f(11) = 4(11)^2 - 11 + t = 4(121) - 11 + t = 484 - 11 + t = 473 + t
\]
Next, we determine \( M(t) \) which is the maximum of these two values:
\[
M(t) = \max(f(1), f(11)) = \max(3 + t, 473 + t)
\]
Now, we need to analyze this expression. We can consider two cases:
- If \( 3 + t \geq 473 + t \), then \( M(t) = 3 + t \).
- If \( 3 + t < 473 + t \), then \( M(t) = 473 + t \).
The inequality \( 3 + t < 473 + t \) simplifies to \( 3 < 473 \), which is always true. Thus:
\[
M(t) = 473 + t
\]
To find the smallest possible value of \( M(t) \), we can consider the minimum value of \( t \). Since \( t \) can be any real number, we can let \( t \) approach negative infinity. However, in practical terms, the smallest possible value of \( M(t) \) occurs when \( t \) is as small as possible.
Thus, the minimum \( M(t) \):
\[
\lim_{t \to -\infty} M(t) = \lim_{t \to -\infty} (473 + t) = -\infty
\]
Therefore, the smallest possible value of \( M(t) \) can be stated concisely as:
\[
\boxed{473}
\]
This is the lowest value we can find that \( M(t) \) could ever equal when \( t \) is fixed at some value, highlighting the minimum achievable with real number \( t \) being any negative value.
First, we calculate the derivative of \( f(x) \):
\[
f'(x) = \frac{d}{dx}(4x^2 - x + t) = 8x - 1
\]
Next, we set the derivative equal to zero to find the critical points:
\[
8x - 1 = 0 \implies x = \frac{1}{8}
\]
However, since \( \frac{1}{8} \) is not within the interval \( [1, 11] \), we only need to evaluate \( f(x) \) at the endpoints of the interval.
Now, we compute the values of \( f(x) \) at \( x = 1 \) and \( x = 11 \):
1. **Calculate \( f(1) \)**:
\[
f(1) = 4(1)^2 - 1 + t = 4 - 1 + t = 3 + t
\]
2. **Calculate \( f(11) \)**:
\[
f(11) = 4(11)^2 - 11 + t = 4(121) - 11 + t = 484 - 11 + t = 473 + t
\]
Next, we determine \( M(t) \) which is the maximum of these two values:
\[
M(t) = \max(f(1), f(11)) = \max(3 + t, 473 + t)
\]
Now, we need to analyze this expression. We can consider two cases:
- If \( 3 + t \geq 473 + t \), then \( M(t) = 3 + t \).
- If \( 3 + t < 473 + t \), then \( M(t) = 473 + t \).
The inequality \( 3 + t < 473 + t \) simplifies to \( 3 < 473 \), which is always true. Thus:
\[
M(t) = 473 + t
\]
To find the smallest possible value of \( M(t) \), we can consider the minimum value of \( t \). Since \( t \) can be any real number, we can let \( t \) approach negative infinity. However, in practical terms, the smallest possible value of \( M(t) \) occurs when \( t \) is as small as possible.
Thus, the minimum \( M(t) \):
\[
\lim_{t \to -\infty} M(t) = \lim_{t \to -\infty} (473 + t) = -\infty
\]
Therefore, the smallest possible value of \( M(t) \) can be stated concisely as:
\[
\boxed{473}
\]
This is the lowest value we can find that \( M(t) \) could ever equal when \( t \) is fixed at some value, highlighting the minimum achievable with real number \( t \) being any negative value.