question consider the function f(x) below. over what interval(s) is the function decreasing and concave…

Explanation:

Step1: Find the first - derivative

First, find the derivative of $f(x)=x^{3}-5x^{2}+21x + 1$ using the power rule. The power rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. So, $f^\prime(x)=3x^{2}-10x + 21$.

Step2: Find the second - derivative

Differentiate $f^\prime(x)$ to get the second - derivative. $f^{\prime\prime}(x)=6x-10$.

Step3: Determine where the function is decreasing

A function is decreasing when $f^\prime(x)<0$. For $f^\prime(x)=3x^{2}-10x + 21$, the discriminant of the quadratic $ax^{2}+bx + c$ (here $a = 3$, $b=-10$, $c = 21$) is $\Delta=b^{2}-4ac=(-10)^{2}-4\times3\times21=100 - 252=-152<0$ and $a = 3>0$, so $f^\prime(x)>0$ for all real $x$, and the function is never decreasing.

Step4: Determine where the function is concave down

A function is concave down when $f^{\prime\prime}(x)<0$. Set $f^{\prime\prime}(x)=6x - 10<0$. Solving for $x$ gives $6x<10$, or $x<\frac{5}{3}$. In interval notation, the interval is $(-\infty,\frac{5}{3})$. Since the function is never decreasing, and is concave down on $(-\infty,\frac{5}{3})$, and we need the intersection of decreasing and concave - down intervals, the answer is $\varnothing$.

Answer:

$\varnothing$