determine the intervals on which the following function is concave up or concave down. identify any…

determine the intervals on which the following function is concave up or concave down. identify any inflection points. f(x)=x^4 - 4x^3 + 7\na. (simplify your answer. type your answer in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)\nthe function is concave up on (-∞,0),(2,∞) and concave down on (0,2).\nb. (simplify your answers. type your answers in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)\nthe function is concave up on and the function is never concave down.\nc. (simplify your answer. type your answer in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)\nd. the function is never concave up nor concave down.\nlocate any inflection points of f. select the correct choice and, if necessary, fill in the answer box to complete your choice.\na. an inflection point occurs at x=\n(type an integer or a simplified fraction. use a comma to separate answers as needed.)\nb. there are no inflection points for f.
Answer
Explanation:
Step1: Find the first - derivative
Using the power rule $(x^n)'=nx^{n - 1}$, for $f(x)=x^{4}-4x^{3}+7$, we have $f'(x)=4x^{3}-12x^{2}$.
Step2: Find the second - derivative
Differentiate $f'(x)$ again. $f''(x)=(4x^{3}-12x^{2})'=12x^{2}-24x = 12x(x - 2)$.
Step3: Find the intervals of concavity
Set $f''(x)=0$, then $12x(x - 2)=0$. The solutions are $x = 0$ and $x = 2$. Test the intervals:
- For $x\lt0$, let $x=-1$. Then $f''(-1)=12\times(-1)\times(-1 - 2)=36\gt0$, so the function is concave up on $(-\infty,0)$.
- For $0\lt x\lt2$, let $x = 1$. Then $f''(1)=12\times1\times(1 - 2)=-12\lt0$, so the function is concave down on $(0,2)$.
- For $x\gt2$, let $x = 3$. Then $f''(3)=12\times3\times(3 - 2)=36\gt0$, so the function is concave up on $(2,\infty)$.
Step4: Find the inflection points
Since the concavity changes at $x = 0$ and $x = 2$, these are the inflection points.
Answer:
The function is concave up on $(-\infty,0),(2,\infty)$ and concave down on $(0,2)$. An inflection point occurs at $x = 0,2$. So for the first part, the answer is A. For the second part, the answer is A. 0,2.