Here's the question you clicked on:
malibugranprix2000
solving an equation If f(x)=x^3-8x+10, show that there are values c for which f(c) equals; a)pi b)-square root of 3 c)5,000,000
for each part proceed this way... a) pi let there be a number c such that f(c) = pi i.e x^3 - 8x + 10 = pi x^3 -8x +10-pi = 0 all you need to show is that there is at least one real root of this equation and you are done
so for b is -9.05
oh I plugged something wrong. Is it, 18.66
10 + sqrt(3) = 11.732 x^3 -8x +11.732 = 0 for 2nd part
\(f(x)\) is continuous, so it can possess any value between \(f(a)\) and \(f(b)\). If you put \(a=-\infty\) and \(b=\infty\), you will have, that your polynomial can possess any value from \(\mathbb R\). This is the proof.
Basically, you want to use the intermediate value theorem.