Example

Sketch the graph of \(y=x^3+x^2-8x-12\).

Solution

1. We first find the \(x\)-intercepts and \(y\)-intercepts.
   When \(x=0\), we have \(y=-12\), and so the \(y\)-intercept is \(-12\).
   When \(y=0\), \begin{align*} 0 &= x^3+x^2-8x-12 \\ &=(x+2)^2(x-3), \end{align*} so \(x=-2\) or \(x=3\). The \(x\)-intercepts are 3 and \(-2\).
2. The derivative is \[ \dfrac{dy}{dx} = 3x^2+2x-8 = (3x-4)(x+2). \] So \(\dfrac{dy}{dx}=0\) implies \(x=-2\) or \(x=\dfrac{4}{3}\).
   The coordinates of the stationary points are \((-2,0)\) and \((\dfrac{4}{3},-\dfrac{500}{27})\).
3. The graph of \(\dfrac{dy}{dx}\) against \(x\) is a parabola with a positive coefficient of \(x^2\). We can see:

• \(\dfrac{dy}{dx} = 0\) if and only if \(x=-2\) or \(x=\dfrac{4}{3}\)
• \(\dfrac{dy}{dx} > 0\) if and only if \(x<-2\) or \(x>\dfrac{4}{3}\)
• \(\dfrac{dy}{dx} < 0\) if and only if \(-2<x<\dfrac{4}{3}\).

So the graph is increasing on \((-\infty,-2) \cup (\dfrac{4}{3},\infty)\) and decreasing on \((-2,\dfrac{4}{3})\).

Value of \(x\)		−2		\(\dfrac{4}{3}\)
Sign of \(\dfrac{dy}{dx}\)	+	0	−	0	+
Slope of graph	\(\diagup\)	—	\(\diagdown\)	—	\(\diagup\)

4. From the investigation of the sign of \(\dfrac{dy}{dx}\), we see that there is a local maximum at \(x = -2\), and a local minimum at \(x = \dfrac{4}{3}\).
5. As \(x \to \infty\), \(y \to \infty\), and as \(x \to -\infty\), \(y \to -\infty\).
6. The second derivative is \[ \dfrac{d^2 y}{dx^2} = 6x+2. \] If \(x < -\dfrac{1}{3}\), then \(\smash{\dfrac{d^2 y}{dx^2}} < 0\), and if \(x > -\dfrac{1}{3}\), then \(\smash{\dfrac{d^2 y}{dx^2}} > 0\). There is an inflexion point where \(x = -\dfrac{1}{3}\).

Example

Sketch the graph of \(y = \sqrt{x} - x^2\), for \(x\geq 0\).

Solution

The first and second derivatives are

\[ \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} - 2x,\qquad \dfrac{d^2y}{dx^2} = -\dfrac{1}{4}x^{-\frac{3}{2}} - 2, \qquad\text{for } x>0. \]

We follow the six steps for sketching polynomial graphs, which are still useful for this type of graph.

1. We first find the \(x\)-intercepts: \begin{align*} \sqrt{x} - x^2 &= 0 \\ \sqrt{x}(1-x^{\frac{3}{2}}) &= 0, \end{align*} which gives \(x = 0\) or \(x = 1\).
2. We find the stationary points by solving \(\dfrac{dy}{dx} = 0\): \begin{align*} \dfrac{1}{2\sqrt{x}} - 2x &= 0 \\ 1 &= 4x^\frac{3}{2} \\ x^\frac{3}{2} &= \dfrac{1}{4} \\ x &= \Bigl(\dfrac{1}{4}\Bigr)^{\frac{2}{3}}. \end{align*} So there is a stationary point at \(x = \bigl(\dfrac{1}{4}\bigr)^{\frac{2}{3}} = 2^{-\frac{4}{3}}\).
3. If \(0 < x < 2^{-\frac{4}{3}}\), then \(\dfrac{dy}{dx} > 0\), and if \(x > 2^{-\frac{4}{3}}\), then \(\dfrac{dy}{dx} < 0\).

   We can summarise this in a gradient diagram.

   Value of \(x\)		\(2^{-\frac{4}{3}}\)
   Sign of \(\dfrac{dy}{dx}\)	+	0	−
   Slope of graph	\(\diagup\)	—	\(\diagdown\)

   Hence, there is a local maximum at \(x = 2^{-\frac{4}{3}}\). The corresponding \(y\)-value is \begin{align*} y &= \bigl(2^{-\frac{4}{3}}\bigr)^{\frac{1}{2}} - \bigl(2^{-\frac{4}{3}}\bigr)^2 = 2^{-\frac{2}{3}} - 2^{-\frac{8}{3}} \\ &= 2^{-\frac{8}{3}}\bigl(2^2 - 1\bigr) = 3 \times 2^{-\frac{8}{3}}. \end{align*}
4. We can also determine that there is a local maximum at \(x = 2^{-\frac{4}{3}}\) by carrying out the second derivative test. When \(x = 2^{-\frac{4}{3}}\), \[ \dfrac{d^2y}{dx^2} = -\dfrac{1}{4}\bigl(2^{-\frac{4}{3}}\bigr)^{-\frac{3}{2}} - 2 = -\dfrac{1}{4}\bigl(2^2\bigr) - 2 = -3. \] Since \(\dfrac{dy}{dx} = 0\) and \(\dfrac{d^2y}{dx^2} < 0\) at \(x = 2^{-\frac{4}{3}}\), the point \(x = 2^{-\frac{4}{3}}\) is a local maximum.
5. As \(x \to \infty\), \(y \to -\infty\). We do not consider \(x \to -\infty\), since the domain of the function is \([0,\infty)\).
6. For all \(x > 0\), we have \[ \dfrac{d^2y}{dx^2} = -\dfrac{1}{4}x^{-\frac{3}{2}} - 2 < 0. \]

   Hence, there are no points of inflexion.

We can now sketch the graph.

Example

Sketch the graph of \(f(x) = e^x+e^{-2x}\).

Solution

The domain of \(f\) is all real numbers. The first two derivatives of \(f\) are \begin{align*} f'(x) &= e^x-2e^{-2x} \\ f''(x) &= e^x+4e^{-2x}. \end{align*}

1. For all \(x\), we have \(e^x > 0\) and \(e^{-2x} > 0\), and therefore \(f(x) > 0\). Hence, there are no \(x\)-intercepts. We have \(f(0) = 2\), and so the \(y\)-intercept is 2.
2. We find the stationary points by solving \(f'(x) = 0\): \begin{align*} e^x - 2e^{-2x} &= 0 \\ e^{3x} &= 2 \\ x &= \dfrac{1}{3}\log_e 2. \end{align*} So there is a stationary point at \(x = \dfrac{1}{3}\log_e 2\). The value of the function at this point is \(f(\dfrac{1}{3}\log_e 2) = 3 \times 2^{-\frac{2}{3}}\).
3. If \(x<\dfrac{1}{3}\log_e 2\), the gradient is negative and the graph of \(y=f(x)\) is decreasing.
   If \(x>\dfrac{1}{3}\log_e 2\), the gradient is positive and the graph of \(y=f(x)\) is increasing.
4. We have \(f''(x) = e^x+4e^{-2x} > 0\), for all \(x\). So there is a local minimum at \(x=\dfrac{1}{3}\log_e 2\).
5. As \(x \to \infty\), \(f(x) \to \infty\), and as \(x \to -\infty\), \(f(x) \to \infty\).
6. Since \(f''(x) = e^x+4e^{-2x} > 0\), for all \(x\), there are no inflexion points.

Example

Sketch the graph of \(f(x) = e^{-x}+x\).

Solution

The domain of \(f\) is all real numbers. The first two derivatives are \(f'(x) = -e^{-x}+1\) and \(f''(x) = e^{-x}\).

1. We have \(f(0) = 1\), so the \(y\)-intercept is 1. We will see later that there is a local minimum at \(x=0\). Indeed, the minimum value of the function occurs when \(x=0\). Hence, there are no \(x\)-intercepts.
2. Solving \(f'(x) = 0\) gives \[ -e^{-x}+1 = 0 \ \iff\ e^{-x} = 1 \ \iff\ x = 0. \] Hence, there is a stationary point at \(x =0\).
3. If \(x<0\), the gradient is negative and the graph of \(y=f(x)\) is decreasing.
   If \(x>0\), the gradient is positive and the graph of \(y= f(x)\) is increasing.
4. We have \(f''(x) = e^{-x} > 0\), for all \(x\). Hence, there is a local minimum when \(x=0\).
5. If \(x \to -\infty\), then \(f(x) \to \infty\). If \(x \to \infty\), then the line \(y=x\) is an asymptote, since \(e^{-x} \to 0\).

Example

Sketch the graph of the function \(f \colon [1,3] \to \mathbb{R}\) given by \(f(x) = x^2(x-4)\), and find the maximum and minimum values of the function.

Solution

If \(f(x) = 0\), then \(x=0\) or \(x=4\), which are outside the domain. Therefore the graph has no \(x\)-intercepts.

We next find the derivative and the second derivative of \(f\):

\begin{align*} f'(x) &= 3x^2-8x = x(3x-8) \\ f''(x) &= 6x-8. \end{align*}

Hence, there is a stationary point at \(x=\dfrac{8}{3}\), and a point of inflexion at \(x=\dfrac{4}{3}\).

To determine if \(x=\dfrac{8}{3}\) is a local maximum or minimum, we use the second derivative test:

\[ f''\Bigl(\dfrac{8}{3}\Bigr) = 8 > 0, \]

and hence there is a local minimum at \(x=\dfrac{8}{3}\).

We now find the values of \(f\) at the two endpoints and the local minimum:

\[ f\Bigl(\dfrac{8}{3}\Bigr) = -\dfrac{256}{27},\qquad f(1) = -3,\qquad f(3) = -9. \]

Hence, the maximum value of \(f\) is \(-3\), and it occurs when \(x=1\). The minimum value of \(f\) is \(-\dfrac{256}{27}\), and it occurs when \(x=\dfrac{8}{3}\).

Example

Define \(f \colon [-7,6] \to \mathbb{R}\) by \(f(x) = 3x^4+8x^3-174x^2-360x\). Find the maximum and minimum values of the function.

Solution

The first and second derivatives are

\begin{align*} f'(x) &= 12x^3+24x^2-348x-360 \\ &= 12(x^3+2x^2-29x-30) = 12(x+1)(x-5)(x+6) \\ f''(x) &= 36x^2+48x-348 = 12(3x^2+4x-29). \end{align*}

So \(f'(x)=0\) implies \(x = 5\) or \(x = -1\) or \(x=-6\). There are stationary points at these three values. We use the second derivative test:

\[ f''(5) = 792 > 0,\qquad f''(-1) = -360 < 0,\qquad f''(-6) = 660 > 0. \]

Hence, there is a local minimum at \(x=5\), a local maximum at \(x=-1\), and a local minimum at \(x=-6\).

We have the following points on the graph:

• the left endpoint \((-7,-1547)\)
• a local minimum \((-6,-1944)\)
• a local maximum \((-1,181)\)
• a local minimum \((5,-3275)\)
• the right endpoint \((6,-2808)\).

The maximum value of the function \(f\) is \(181\), which occurs at \(x=-1\), and the minimum value of \(f\) is \(-3275\), which occurs at \(x=5\).