What are the extrema of f(x)=5x^2+4x-3 on [-oo,oo]?

Aug 2, 2018

Local minimum: $\left(- \frac{2}{5} , - \frac{19}{5}\right)$

Explanation:

Local extrema occur when the first derivative returns a zero whereas the second derivative returns a non-zero value.

Apply the Power Rule to find the first and second derivative of $f \left(x\right)$

$f ' \left(x\right) = \left(2 \times 5\right) \textcolor{w h i t e}{l} x + 4 = 10 \textcolor{w h i t e}{l} x + 4$
$f ' ' \left(x\right) = 10$

Set $f ' \left(x\right)$ to zero and solve for $x$:

$10 \textcolor{w h i t e}{l} x + 4 = f ' \left(x\right) = 0$
$x = - \frac{2}{5}$

The solution to $f ' \left(x\right) = 0$ is unique, meaning that $f \left(x\right)$ has only a single local extremum on $x \in \left(- \infty , + \infty\right)$.

Evaluate the second derivative at $x = - \frac{2}{5}$:

$f ' ' \left(- \frac{2}{5}\right) = 10 > 0$

$f ' \left(x\right) = 0$ and $f ' ' \left(x\right)$ is positive at $x = - \frac{2}{5}$. Therefore $f \left(x\right)$ is concave upward and has a local minimum with coordinates $\left(- \frac{2}{5} , f \left(- \frac{2}{5}\right)\right)$.

$f \left(- \frac{2}{5}\right) = - \frac{19}{5} = - 3.8$. Therefore $f \left(x\right)$ has a local minimum at $\left(- \frac{2}{5} , - \frac{19}{5}\right)$ or equivalently $\left(- 0.4 , - 3.8\right)$ as seen on the plot:

graph{5 x ^2 + 4x -3 [-2, 1, -5.567, 0.676]}