if $f(x)=x^{3}-4x + 6$, show that there are values $c$ for which $f(c)$ equals (a) $pi$, (b) $-sqrt{3}$, and…

if $f(x)=x^{3}-4x + 6$, show that there are values $c$ for which $f(c)$ equals (a) $pi$, (b) $-sqrt{3}$, and (c) $5000000$. now, find $f(170)$. $f(170)=4912326$ (simplify your answer.) does a solution exist between 165 and 170 for $f(x)=5000000$? inconclusive, because $5000000$ does not lie between $f(165)$ and $f(170)$. yes, because $f(165)>5000000>f(170)$. inconclusive, because $f(165)<0$ but $f(170)>0$. yes, because $f(165)<5000000<f(170)$.

if $f(x)=x^{3}-4x + 6$, show that there are values $c$ for which $f(c)$ equals (a) $pi$, (b) $-sqrt{3}$, and (c) $5000000$. now, find $f(170)$. $f(170)=4912326$ (simplify your answer.) does a solution exist between 165 and 170 for $f(x)=5000000$? inconclusive, because $5000000$ does not lie between $f(165)$ and $f(170)$. yes, because $f(165)>5000000>f(170)$. inconclusive, because $f(165)<0$ but $f(170)>0$. yes, because $f(165)<5000000<f(170)$.

Answer

Explanation:

Step1: Recall Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function $y = f(x)$ is continuous on a closed interval $[a,b]$, and $k$ is a number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a,b)$ such that $f(c)=k$.

Step2: Analyze the given function

The function $f(x)=x^{3}-4x + 6$ is a polynomial function. Polynomial functions are continuous everywhere, so $f(x)$ is continuous on any closed interval $[a,b]$.

Step3: Evaluate the condition for $k = 5000000$

We know that $f(165)=(165)^{3}-4\times165 + 6=165^{3}-660 + 6=4492125-660 + 6=4491471$ and $f(170)=4912326$. Since $f(165)=4491471<5000000<f(170)=4912326$, by the Intermediate - Value Theorem, there exists a number $c$ in the interval $(165,170)$ such that $f(c)=5000000$.

Answer:

Yes, because $f(165)<5,000,000<f(170)$