Answer:
Relative minimum: [tex]\left(-\frac{5}{2}, -\frac{33}{4}\right)[/tex], Relative maximum: [tex]DNE[/tex]
Step-by-step explanation:
First, we obtain the First and Second Derivatives of the polynomic function:
First Derivative
[tex]f'(x) = 2\cdot x + 5[/tex] (1)
Second Derivative
[tex]f''(x) = 2[/tex] (2)
Now, we proceed with the First Derivative Test on (1):
[tex]2\cdot x + 5 = 0[/tex]
[tex]x = -\frac{5}{2}[/tex]
The critical point is [tex]-\frac{5}{2}[/tex].
As the second derivative is a constant function, we know that critical point leads to a minimum by Second Derivative Test, since [tex]f\left(-\frac{5}{2}\right) > 0[/tex].
Lastly, we find the remaining component associated with the critical point by direct evaluation of the function:
[tex]f\left(-\frac{5}{2} \right) = \left(-\frac{5}{2} \right)^{2} + 5\cdot \left(-\frac{5}{2} \right) - 2[/tex]
[tex]f\left(-\frac{5}{2} \right) = -\frac{33}{4}[/tex]
There are relative maxima.
