Question

Find all relative extrema of the function. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE.)
f(x) = x² + 5x – 2

relative maximum
(x, y) = DNE


relativo minimum
(x, y) =

Answer 1

Answer:

Relative minimum: $\left(-\frac{5}{2}, -\frac{33}{4}\right)$, Relative maximum: $DNE$

Step-by-step explanation:

First, we obtain the First and Second Derivatives of the polynomic function:

First Derivative

$f'(x) = 2\cdot x + 5$ (1)

Second Derivative

$f''(x) = 2$ (2)

Now, we proceed with the First Derivative Test on (1):

$2\cdot x + 5 = 0$

$x = -\frac{5}{2}$

The critical point is $-\frac{5}{2}$.

As the second derivative is a constant function, we know that critical point leads to a minimum by Second Derivative Test, since $f\left(-\frac{5}{2}\right) > 0$.

Lastly, we find the remaining component associated with the critical point by direct evaluation of the function:

$f\left(-\frac{5}{2} \right) = \left(-\frac{5}{2} \right)^{2} + 5\cdot \left(-\frac{5}{2} \right) - 2$

$f\left(-\frac{5}{2} \right) = -\frac{33}{4}$

There are relative maxima.