Question
Find the maximum and minimum values, if any, of the function given by:
$f(x) = -(x - 1)^2 + 10$
$f(x) = -(x - 1)^2 + 10$
✓
Answer
Given: $\text{f}\text{(x)} = - (\text{x} - 2)^2 + 10\ \dots\text{(i)}$
$\text{Since } \text{(x} - 1)^2\geq0\text{ for all }\text{x}\in \text{R}$
Multiplying both sides by $-1$ and adding $10$ both sides,
$-(\text{x} - 1)^2 + 10\leq10\ \Rightarrow\ \text{f}\text{(x)} \leq10\ \ [\text{using eq. (i)}]$
Therefore, maximum value of $f(x)$ is $10$ which is obtained when $x - 1 = 0$ i. e., $x = 1$
And therefore, minimum value of $f(x)$ does not exist.
$\text{Since } \text{(x} - 1)^2\geq0\text{ for all }\text{x}\in \text{R}$
Multiplying both sides by $-1$ and adding $10$ both sides,
$-(\text{x} - 1)^2 + 10\leq10\ \Rightarrow\ \text{f}\text{(x)} \leq10\ \ [\text{using eq. (i)}]$
Therefore, maximum value of $f(x)$ is $10$ which is obtained when $x - 1 = 0$ i. e., $x = 1$
And therefore, minimum value of $f(x)$ does not exist.
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