Q. 63 - Calculus

Question

Sketch careful, labeled graphs of each function f in Exercises 63–82 by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f, f\text{'}, and f\text{'}\text{'},$ and examine any relevant limits so that you can describe all key points and behaviors of f.

$f\left(x\right)=x^2+3x$

Step-by-Step Solution

Answer

The sign chart is

The sketch of the graph is

Step 1. Given Information.

The given function is $f\left(x\right)=x^2+3x.$

Step 2. Sketch the labeled graph.

To sketch the labeled graph of f, we will use theorem 3.6 and 3.10.

Theorem 3.6 states that the Derivative Measures Where a Function is Increasing or Decreasing, let f be a function that is differentiable on an interval I.

(a) If $f\text{'}$ is positive in the interior of I, then f is increasing on I.

(b) If $f\text{'}$ is negative in the interior of I, then f is decreasing on I.

(c) If $f\text{'}$ is zero in the interior of I, then f is constant on I.

Theorem 3.10 states that the Second Derivative Determines Concavity, suppose both f and $f\text{'}$ are differentiable on an interval I.

(a) If $f\text{'}\text{'}$ is positive on I, then f is concave up on I.

(b) If $f\text{'}\text{'}$ is negative on I, then f is concave down on I.

Step 3. Finding the roots.

To find the roots we will put the given function equal to zero.

So,

$$\begin{gathered} f\left(x\right)=x^2+3x \\ 0=x\left(x+3\right) \\ x=0 and x+3=0 \\ x=-3 \end{gathered}$$

Therefore, the given function have roots at $x=0,-3.$

Step 4. Testing the signs of f.

To sketch the sign chart, let's test the signs on both sides.

For f

$$\begin{gathered} f(-4)={\left(-4\right)}^2+3\left(-4\right) \\ f(-4)=4 \\ \mathrm{Now}, f(-1)={(-1)}^2+3(-1) \\ f(-1)=-2 \\ \mathrm{Now}, f\left(1\right)={\left(1\right)}^2+3\left(1\right) \\ f\left(1\right)=4 \end{gathered}$$

Step 5. Testing the signs.

Now, let's test the sign for $f\text{'} and f\text{'}\text{'}.$

Let's differentiate the equation to find $f\text{'}.$

So,

  $$\begin{gathered} f\text{'}\left(x\right)=2x+3 \\ 0=2x+3 \\ x=-\frac{3}{2} \end{gathered}$$

Testing the signs on both sides,

$$\begin{gathered} f\text{'}\left(-3\right)=2\left(-3\right)+3 \\ f\text{'}(-3)=-3 \\ And \\ f\text{'}\left(0\right)=2\left(0\right)+3 \\ f\text{'}\left(0\right)=3 \end{gathered}$$

Thus, $f\text{'}$ is negative on the interval $\left(-\infty ,-\frac{3}{2}\right)$ and positive on the interval $\left(-\frac{3}{2},\infty \right).$ Hence the graph of f will be increasing on the positive intervals and decreasing on the negative intervals.

Let's differentiate again.

So, $f\text{'}\text{'}\left(x\right)=2$

Thus, $f\text{'}\text{'}\left(x\right)$ is positive. Hence f will be concave up everyplace.

Step 6. Sketch the sign chart.

The sign chart is

Step 7. Examine the relevant limit.

Let's examine the limits of 

$f\left(x\right)=x^2+3x \mathrm{as} x\to \pm \infty .$

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\infty \\ \underset{x\to -\infty }{\lim }f\left(x\right)=\infty \end{gathered}$$

Step 8. Sketch the graph of function f.

The graph of the function is