Question

question 2
given ( f(x) = 2x^3 + 12x^2 + 5x + 3 )
find where ( f(x) = 0 )
answer: ( x = ) -2.00 (round to two decimal places if necessary)
at ( x = 0.5 ) the function select an answer
at ( x = 6.5 ) the function select an answer
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Explanation:

Step1: Find first derivative

$f'(x) = \frac{d}{dx}(2x^3 + 12x^2 + 5x + 3) = 6x^2 + 24x + 5$

Step2: Find second derivative

$f''(x) = \frac{d}{dx}(6x^2 + 24x + 5) = 12x + 24$

Step3: Solve $f''(x)=0$

$12x + 24 = 0$
$12x = -24$
$x = -2$

Step4: Test concavity at $x=0.5$

Substitute $x=0.5$ into $f''(x)$:
$f''(0.5) = 12(0.5) + 24 = 6 + 24 = 30 > 0$

Step5: Test concavity at $x=6.5$

Substitute $x=6.5$ into $f''(x)$:
$f''(6.5) = 12(6.5) + 24 = 78 + 24 = 102 > 0$

Answer:

$x = -2.00$
At $x = 0.5$ the function is concave up
At $x = 6.5$ the function is concave up