Asked by glenn
The function f(x) = x2 + 1 has a minimum at what coordinate point (x, f(x))?
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Answered by
GPT-4o mini
The function \( f(x) = x^2 + 1 \) is a quadratic function that opens upwards (since the coefficient of \( x^2 \) is positive). The minimum value of a quadratic function in the form \( ax^2 + bx + c \) occurs at the vertex, which can be found using the formula:
\[
x = -\frac{b}{2a}
\]
For the function \( f(x) = x^2 + 1 \):
- \( a = 1 \)
- \( b = 0 \)
- \( c = 1 \)
Plugging in the values:
\[
x = -\frac{0}{2 \cdot 1} = 0
\]
Next, we find \( f(0) \):
\[
f(0) = (0)^2 + 1 = 1
\]
Therefore, the coordinate point at which the function has its minimum is:
\[
(0, 1)
\]
So, the minimum occurs at the point \( (0, 1) \).
\[
x = -\frac{b}{2a}
\]
For the function \( f(x) = x^2 + 1 \):
- \( a = 1 \)
- \( b = 0 \)
- \( c = 1 \)
Plugging in the values:
\[
x = -\frac{0}{2 \cdot 1} = 0
\]
Next, we find \( f(0) \):
\[
f(0) = (0)^2 + 1 = 1
\]
Therefore, the coordinate point at which the function has its minimum is:
\[
(0, 1)
\]
So, the minimum occurs at the point \( (0, 1) \).
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