Q.Let $R = \{(a, b) : a, b \in \mathbb{Z} \text{ and } b = 2a - 4\}$. If $(a, -2) \in R$ and $(4, b^2) \in R$. Then, write the values of $a$ and $b$.

Substituting $1/x$ into $f$ gives the negative of $f(x)$, so the sum is 0.

Given $f(x) = x^3 - \dfrac{1}{x^3}$. Substitute $\dfrac{1}{x}$ for $x$:

$f\left(\dfrac{1}{x}\right) = \left(\dfrac{1}{x}\right)^3 - \dfrac{1}{\left(\frac{1}{x}\right)^3} = \dfrac{1}{x^3} - x^3$

Now add: …

Honesty note on the stem: as given, the stem reads literally "$f(x) = x^2+3 = 19$", which is not a well-formed single equation -- it looks like a paper/transcription typo that ran together the function definition $f(x)=x^2+3$ and the condition $f(x)=19$ used to ask for the pre-image. We answer under the most natural reading consistent with the question asking for "the pre-image": $f: R \to R$, $f(x) = x^2+3$, find $x$ such that $f(x) = 19$.

Solving:

$x^2 + 3 = 19$

$x^2 = 16$

$x = \pm 4$

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Set $f(x)=19$ and solve $x^2+3=19$.

$x^2+3=19 \implies x^2=16 \implies x=\pm4$ …

Substitute a new variable for the function's argument to rewrite the rule purely in terms of that argument.

We are given $f(x+1) = 3x + 5$ and need $f(x)$.

Let $t = x + 1$, so $x = t - 1$. Substitute into the given rule:

$f(t) = 3(t-1) + 5 = 3t - 3 + 5 = 3t + 2$

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Substituting $x=2$ into $f(x)=x^2/2$ gives $f(2)=2$.

Given $f(x) = \dfrac{x^2}{2}$.

Substitute $x = 2$: …

Direct substitution of $x=0$ gives $f(0)=1$.

Given $f(x) = x^2 + 2x + 1$, we simply substitute $x = 0$:

$f(0) = (0)^2 + 2(0) + 1 = 0 + 0 + 1 = 1$

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$f(4)=0$.

With period $2$: $f(4)=f(4-2)=f(2)=f(0)$. For an odd function $f(-x)=-f(x)$; setting $x=0$ gives $f(0)=-f(0)$, so $f(0)=0$.

Therefore $f(4)=0$. ("Add" in the stem is a transcription of "odd.")

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Plug $x=\tfrac12$ directly into the formula for $f(x)$ and simplify.

$f(x) = x^2 - 4x + 1$

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Substitute $x=0$ directly into the polynomial.

$f(x) = x^3 - 3x^2 + 1$

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