Question

question consider the function f(x) below. over what interval(s) is the function decreasing and concave down? give your answer in interval notation. enter ∅ if the interval does not exist. f(x)=x³ - 5x²+21x + 1 sorry, thats incorrect. try again?

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 $f(x)$ is decreasing

A function $y = f(x)$ 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 $x$.

Step4: Determine where $f(x)$ is concave down

A function $y = f(x)$ is concave down when $f^{\prime\prime}(x)<0$. Set $f^{\prime\prime}(x)=6x - 10<0$. Solving for $x$ gives $6x<10$, or $x<\frac{5}{3}$. Since $f^\prime(x)>0$ for all $x$, there is no interval where the function is both decreasing and concave down.

Answer:

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