Question

hw 20 - absolute extrema section 3.6: problem 8
(1 point)
find the absolute maximum and minimum values of ( f(x)=4 x-9+2 x^{2} ), if any, over the interval ( (-infty,+infty) ).
absolute maximum is and it occurs at ( x= )
absolute minimum is and it occurs at ( x= )
(round to three decimal places as needed.)
notes: if there is more than one ( x ) value, enter as a comma separated list. enter
one\ in any unused answer box.
note: you can earn partial credit on this problem.
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Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=2x^{2}+4x - 9\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=(2x^{2}+4x - 9)^\prime=4x + 4\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(4x+4 = 0\). Solving for \(x\):

$$
LATEXBLOCK0
$$

Step3: Find the second - derivative

Differentiate \(f^\prime(x)=4x + 4\) to get \(f^{\prime\prime}(x)=4\). Since \(f^{\prime\prime}(x)=4>0\), the function \(f(x)\) has a local minimum at \(x =-1\).

Step4: Find the value of the function at the critical point

Substitute \(x=-1\) into \(f(x)\):

$$
LATEXBLOCK1
$$

As \(x\to\pm\infty\), \(y = f(x)=2x^{2}+4x - 9\to+\infty\) (because the leading coefficient of the quadratic function \(y = 2x^{2}+4x - 9\) is \(a = 2>0\)).

Answer:

absolute maximum is none and it occurs at \(x=\) none; absolute minimum is \(-11\) and it occurs at \(x=-1\)