Calculus and Its Applications, Chapter 1, 1.5, Section 1.5, Problem 92

If $f(x) = x^2 + 7x + 2$. Determine the interval(s) for which $f'(x)$ is positive.
We are seeking the interval(s) at which $f'(x) > 0$. So,

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{d}{dx} (x^2 + 7x + 2)\\
\\
&= \frac{d}{dx} (x^2) + \frac{d}{dx} (7x) + \frac{d}{dx} (2)\\
\\
&= \frac{d}{dx} (x^2) + 7 \cdot \frac{d}{dx} (x) + \frac{d}{dx} (2)\\
\\
&= 2x^{2- 1} + 7 \cdot (1) + 0 \\
\\
&= 2x + 7
\end{aligned}
\end{equation}
$


Thus, we have

$
\begin{equation}
\begin{aligned}
2x + 7 &> 0 \\
\\
2x &> - 7 \\
\\
x &> - \frac{7}{2}
\end{aligned}
\end{equation}
$

Therefore, the interval for which $f'(x)$ is positive is $\displaystyle \left( -\frac{7}{2}, \infty \right)$