find all relative extrema of the function. use the second derivative test where applicable. (if an answer…

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)\n\nf(x) = x² + 9x - 3\n\nrelative maximum (x, y) = (\n\nrelative minimum (x, y) = (\n

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)\n\nf(x) = x² + 9x - 3\n\nrelative maximum (x, y) = (\n\nrelative minimum (x, y) = (\n

Answer

Explanation:

Step1: Find the first derivative

The function is ( f(x)=x^{2}+9x - 3 ). Using the power rule ((x^n)^\prime=nx^{n - 1}), the first derivative (f^\prime(x)=2x + 9).

Step2: Find the critical points

Set (f^\prime(x) = 0), so (2x+9 = 0). Solving for (x), we get (x=-\frac{9}{2}).

Step3: Find the second derivative

Differentiate (f^\prime(x)=2x + 9) with respect to (x). Using the power rule, (f^{\prime\prime}(x)=2).

Step4: Apply the second - derivative test

Since (f^{\prime\prime}(x)=2>0) for all (x), the function (f(x)) is concave up at (x =-\frac{9}{2}).

Now, find the (y) - value. Substitute (x =-\frac{9}{2}) into (f(x)): [ \begin{align*} f\left(-\frac{9}{2}\right)&=\left(-\frac{9}{2}\right)^{2}+9\left(-\frac{9}{2}\right)-3\ &=\frac{81}{4}-\frac{81}{2}-3\ &=\frac{81 - 162 - 12}{4}\ &=\frac{81-(162 + 12)}{4}\ &=\frac{81 - 174}{4}\ &=-\frac{93}{4} \end{align*} ]

Answer:

relative maximum ((x,y)=\text{DNE})

relative minimum ((x,y)=\left(-\frac{9}{2},-\frac{93}{4}\right))