The absolute minimum of $f ( x ) = \frac { 9 } { x } + x - 3$ on the interval 1 $\le$ x$\le$9 is 1.

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To find the absolute minimum of $f(x) = \frac{9}{x} + x - 3$ on the interval $1 \leq x \leq 9$, we first find the derivative of $f(x)$, set it equal to zero, and solve for $x$ to find critical points. The derivative $f'(x) = -\frac{9}{x^2} + 1$. Setting $f'(x) = 0$ gives $x = 3$. Evaluating $f(x)$ at the endpoints and at $x = 3$, we find $f(1) = 7$, $f(9) = \frac{25}{3}$, and $f(3) = 3$. Therefore, the absolute minimum value of $f(x)$ on the interval is 3, not 1.

The Absolute Minimum Of f(x)=9x+x−3f ( x ) = \frac { 9 } { x } + x - 3f(x)=x9​+x−3

The Absolute Minimum Of $f ( x ) = \frac { 9 } { x } + x - 3$