Question Number 179025 by jlewis last updated on 23/Oct/22
$$\mathrm{2}^{\mathrm{10}{x}} −{x}^{\mathrm{5}} −\mathrm{4}=\mathrm{0} \\ $$
Commented by MJS_new last updated on 23/Oct/22
$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{31950087530} \\ $$$${x}_{\mathrm{2}} \approx.\mathrm{200011544430} \\ $$
Commented by jlewis last updated on 24/Oct/22
$${how}\:{do}\:{u}\:{reach}\:{to}\:{the}\:{approximation}? \\ $$
Commented by MJS_new last updated on 24/Oct/22
![try a few values f(x)=2^(10x) −x^5 −4 f(−2)≈28>0 f(−1)≈−3<0 f(0)=−3<0 f(1)=1019>0 ⇒ we must search in these intervals ]−2; −1[ ]0; 1[](https://www.tinkutara.com/question/Q179079.png)
$$\mathrm{try}\:\mathrm{a}\:\mathrm{few}\:\mathrm{values} \\ $$$${f}\left({x}\right)=\mathrm{2}^{\mathrm{10}{x}} −{x}^{\mathrm{5}} −\mathrm{4} \\ $$$${f}\left(−\mathrm{2}\right)\approx\mathrm{28}>\mathrm{0} \\ $$$${f}\left(−\mathrm{1}\right)\approx−\mathrm{3}<\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{3}<\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1019}>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{must}\:\mathrm{search}\:\mathrm{in}\:\mathrm{these}\:\mathrm{intervals} \\ $$$$\left.\right]−\mathrm{2};\:−\mathrm{1}\left[\right. \\ $$$$\left.\right]\mathrm{0};\:\mathrm{1}\left[\right. \\ $$