Sketch the graphs of $y = \dfrac{1}{f(x)}$, $y = f'(x)$, $y = |f(x)|$, $y = f(|x|)$.

1. Determine the key features and sketch $y = \dfrac{1}{f(x)}$.

• Horizontal Asymptote: As $x \to -\infty$, $f(x) \to 4$, so $y = \dfrac{1}{4}$ is a horizontal asymptote.
• Turning Points: The local maximum at $A(2, 10)$ becomes a local minimum at $(2, 0.1)$.
• Vertical Asymptotes & Roots: The $x$-intercept of $f(x)$ between $x=5$ and $x=6$ becomes a vertical asymptote. As $f(x) \to -\infty$ when $x \to 6^-$, $\dfrac{1}{f(x)} \to 0$, meaning the graph approaches a 'hole' at $(6, 0)$.
• Right Branch: The point $B(12, 1)$ maps to $(12, 1)$. As $x \to \infty$, $f(x) \to 0^+$, so $\dfrac{1}{f(x)} \to +\infty$.

2. Determine the key features and sketch $y = f'(x)$.

• $x$-intercepts: The stationary point at $x=2$ means $f'(2) = 0$, giving an $x$-intercept at $(2, 0)$. Assuming $B(12, 1)$ is a stationary point due to the undulating nature of the curve, there is another $x$-intercept at $(12, 0)$.
• Asymptotes: As $x \to -\infty$ and $x \to \infty$, the graph of $f(x)$ flattens out (approaching $y=4$ and $y=0$ respectively), so $f'(x) \to 0$, giving a horizontal asymptote at $y=0$.
• Vertical Asymptote: As $x \to 6^-$, the curve gets infinitely steep downwards, so $f'(x) \to -\infty$ with a vertical asymptote at $x=6$.
• Signs: $f'(x) > 0$ for $x < 2$, and $f'(x) < 0$ for $2 < x < 6$.

3. Determine the key features and sketch $y = |f(x)|$.

• Reflection: The portion of $f(x)$ below the $x$-axis (between the $x$-intercept and $x=6$) is reflected upwards.
• Kinks: The $x$-intercept between $x=5$ and $x=6$ becomes a sharp point (kink) on the $x$-axis due to the modulus function.
• Asymptote: As $x \to 6^-$, the reflected curve now tends to $+\infty$ instead of $-\infty$.
• Unchanged Parts: The local maximum $A(2, 10)$, the $y=4$ asymptote, and the entire right branch including $B(12, 1)$ remain exactly the same as $f(x) > 0$ in these regions.

4. Determine the key features and sketch $y = f(|x|)$.

• Domain Restriction: Keep the graph of $y = f(x)$ exactly as it is for $x \ge 0$.
• Reflection for $x < 0$: Reflect the $x \ge 0$ portion across the $y$-axis to obtain the graph for $x < 0$. The original left-side asymptote $y=4$ is discarded.
• New Features: By symmetry, a new local maximum appears at $(-2, 10)$, a new vertical asymptote at $x = -6$, and a new point at $(-12, 1)$. The $y$-intercept forms a sharp kink where the two symmetrical halves meet. As $x \to -\infty$, the curve approaches the $x$-axis from above, mirroring the right branch.

1. Extract the key features of the given f(x)

From figure (b): $y=f(x)$ has a vertical asymptote $x=2$ and an oblique asymptote $y=x-1$. There is a branch on each side of $x=2$, with a stationary point on each branch (the curve turns where it leaves/approaches the asymptotes). Note the gradient of $f$ tends to $1$ far from $x=2$ (parallel to $y=x-1$).

2. Sketch y = 1/f(x)

Apply the reciprocal rules:

• Where $f(x)=0$ (the $x$-intercepts of $f$) $\Rightarrow$ vertical asymptotes of $\tfrac1{f(x)}$.
• Where $f(x)\to\pm\infty$ (the line $x=2$ and along $y=x-1$) $\Rightarrow \tfrac1{f(x)}\to 0$, so the curve hugs the $x$-axis there.
• A local maximum of $f$ (above the axis) becomes a local minimum of $\tfrac1{f(x)}$ and vice-versa; the $y$-value is reciprocated.
• $\tfrac1{f(x)}$ keeps the same sign as $f(x)$ in each region.

3. Sketch y = f'(x)

Use the sign and value of the gradient of $f$:

• The vertical asymptote $x=2$ is retained.
• Far from $x=2$, $f$ approaches the line $y=x-1$ (slope $1$), so $f'(x)\to 1$: the line $y=1$ is a horizontal asymptote.
• At each stationary point of $f$, $f'=0$, so $y=f'(x)$ cuts the $x$-axis at those $x$-values.
• Where $f$ is increasing $f'>0$ (above axis); where $f$ is decreasing $f'<0$ (below axis).

4. Sketch y = |f(x)|

Keep all parts of $f$ that lie on or above the $x$-axis; reflect any part below the $x$-axis upward in the $x$-axis. The result lies entirely in $y\ge 0$; sharp corners appear where $f$ crossed the $x$-axis. The vertical asymptote $x=2$ is kept (the branch now goes to $+\infty$ on the side that previously went to $-\infty$); the oblique asymptote $y=x-1$ becomes $y=|x-1|$ for the relevant branch.

5. Sketch y = f(|x|)

Keep the part of $f$ for $x\ge 0$ and reflect it in the $y$-axis to produce the part for $x<0$ (discarding the original $x<0$ part). The resulting curve is symmetric about the $y$-axis: the asymptote $x=2$ is mirrored to give a second vertical asymptote $x=-2$, and the behaviour near $x=2$ is copied near $x=-2$.