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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) =

Sagot :

Xero099idn

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.

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