Answer :
The correct statement is: [tex]\( f \)[/tex] is concave up over the interval [tex]\((- \infty, 0) \cup (\frac{83}{24}, \infty)\).\\[/tex]
To determine the concavity of the function [tex]\( f(x) = -12x^4 + 83x^3 - x + 23 \)[/tex], we need to analyze the second derivative of the function. Let's compute the second derivative, [tex]\( f''(x) \):[/tex]
[tex]\[ f(x) = -12x^4 + 83x^3 - x + 23, \][/tex]
[tex]\[ f'(x) = -48x^3 + 249x^2 - 1, \][/tex]
[tex]\[ f''(x) = -144x^2 + 498x. \][/tex]
To determine the intervals where [tex]\( f(x) \)[/tex] is concave up or concave down, we need to find the critical points of [tex]\( f''(x) \)[/tex]. Setting [tex]\( f''(x) \)[/tex] equal to zero, we have:
[tex]\[ -144x^2 + 498x = 0, \][/tex]
[tex]\[ x(-144x + 498) = 0. \][/tex]
Solving this equation, we find two critical points: [tex]\( x = 0 \) and \( x = \frac{83}{24} \).[/tex]
Now, we can analyze the concavity of [tex]\( f(x) \)[/tex] based on the intervals [tex]\((- \infty, 0)\), \((0, \frac{83}{24})\), and \((\frac{83}{24}, \infty)\):[/tex]
For [tex]\( x < 0 \), \( f''(x) > 0 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is concave up.
For [tex]\( 0 < x < \frac{83}{24} \), \( f''(x) < 0 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is concave down.
For [tex]\( x > \frac{83}{24} \), \( f''(x) > 0 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is concave up.
Therefore, the correct statement is: [tex]\( f \)[/tex] is concave up over the interval [tex]\((- \infty, 0) \cup (\frac{83}{24}, \infty)\).\\[/tex]
To know more about derivative visit -
brainly.com/question/31585278
#SPJ11
