Question

This is matlab questions,please use Julia and give me Julia commands process to slove the questions.

(3) Find an interval [a,b] on which f(x) = x cos(x) and gle) = 7 / 7 satisfies g(x)=f(). Your interval must be in [0,2]. have endpoints accurate to the nearest thousandth, and be as large as possible. Construct a graph showing both fanda on [o, 2] and on which your interval is. visible.

Answer 1

`Hey,

Note: If you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

clc%clears screen
clear all%clears history
close all%closes all files
format long;
f = @(x) x*cos(x)-1/(2*x);
e=1e-6;
a=0.1;
b=1;
ff=@(x) x.*cos(x);
gg=@(x) 1./(2*x);
hold on;
fplot(ff,[0.1,2]);
fplot(gg,[0.1,2]);
fprintf('Interval is [0,%.3f] and [%.3f 2]\n',fzero(f,[0.1,1]),fzero(f,[1.1,1.4]))
legend('f(x)','g(x)');

- 5 x Pradeep G doo G 3 Search Documentation A MATLAB R2018 HOME +s o New Open Save PLOTS la Find Files 3 Compare - Print 3 D - 0 x APPS EDITOR PUBLISH VIEW Insert E. fx - ] Run Section 8 Go To - Comment % %% Breakpoints Run Run and Advance Run Figure 1 Find Indent 2 1 6 - Advance NAVIGATE EDIT BREAKPOINTS File Edit PRADEEP KALRA Documents MATLAB & Editor - C:\Users\PRADEEP KALRA\Documents\MATLAB\saveee.m +1 saveee. m x bisectionmethodbypradeep.m X google RUN Insert View A Tools Desktop 2. Window 3 Help O E C: Users I QX FILE a Current Folder Name - codegen convertFinal directory1 example first googleLinks mathlab mynewdir IIIIIIIIIIII format long; f = @(x) x*cos(x)-1/(2*x); e-le-6; a=0.1; b=1; ff=@ (x) x. *cos(x); gg=@ (x) 1./(2*x); hold on; fplot (ff,[0.1,2]); fplot (99,[0.1,2]); fprintf('Interval is [0,$.3f] and [4.3 legend ('f(x)', 'g(x)'); New folder questions old simple_gui2 1.png 2.png 3.png Details Command Window Interval is [0,0.893] and [1.239 2] Workspace Name fx >> Value 0.1000 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1.0000e-06 @(x)x*cos(x)-1/(2*x) @(x)x.* cos(x) @(x)1./(2*x) 99 script Col 1 In 16 ENG 4:18 PM Type here to search A la US 03/03/20

Kindly revert for any queries

Thanks.