Find the maximum and minimum values of the function given by $h(x) = x + 1, x \in (-1, 1)$.
(NONE) The given function is $h(x) = x + 1$ defined on the open interval $(-1, 1)$.
For any $x \in (-1, 1)$,we have $-1 < x < 1$.
Adding $1$ to all parts of the inequality,we get $-1 + 1 < x + 1 < 1 + 1$,which simplifies to $0 < h(x) < 2$.
As $x$ approaches $-1$ from the right,$h(x)$ approaches $0$,but $h(x)$ is never equal to $0$ because $-1$ is not included in the domain.
As $x$ approaches $1$ from the left,$h(x)$ approaches $2$,but $h(x)$ is never equal to $2$ because $1$ is not included in the domain.
Since the function is strictly increasing and the interval is open,there is no point $c \in (-1, 1)$ such that $h(c)$ is the maximum or minimum value.
Therefore,the function $h(x)$ has neither a maximum nor a minimum value in the interval $(-1, 1)$.
For any $x \in (-1, 1)$,we have $-1 < x < 1$.
Adding $1$ to all parts of the inequality,we get $-1 + 1 < x + 1 < 1 + 1$,which simplifies to $0 < h(x) < 2$.
As $x$ approaches $-1$ from the right,$h(x)$ approaches $0$,but $h(x)$ is never equal to $0$ because $-1$ is not included in the domain.
As $x$ approaches $1$ from the left,$h(x)$ approaches $2$,but $h(x)$ is never equal to $2$ because $1$ is not included in the domain.
Since the function is strictly increasing and the interval is open,there is no point $c \in (-1, 1)$ such that $h(c)$ is the maximum or minimum value.
Therefore,the function $h(x)$ has neither a maximum nor a minimum value in the interval $(-1, 1)$.
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Let $S$ be the set of all twice differentiable functions $f$ from $R$ to $R$ such that $\frac{d^2 f}{d x^2}(x) > 0$ for all $x \in (-1, 1)$. For $f \in S$,let $X_f$ be the number of points $x \in (-1, 1)$ for which $f(x) = x$. Then which of the following statements is(are) true?
$(A)$ There exists a function $f \in S$ such that $X_f = 0$
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$(C)$ There exists a function $f \in S$ such that $X_f = 2$
$(D)$ There does $NOT$ exist any function $f$ in $S$ such that $X_f = 1$
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$(C)$ There exists a function $f \in S$ such that $X_f = 2$
$(D)$ There does $NOT$ exist any function $f$ in $S$ such that $X_f = 1$
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