Question

please explain how to do step 5 in matlab commands.

med at x=c. 2 The first derivative Ne Scr We investigate the function f(x) 4 12x3+9x2. >> x-linspace (-3,3) >> y-41x.^4-12*x.^3 >> plot (x,y), grid 9*x."2; + A plot over the interval I-3,3] reveals an apparent "flat section"' with no visible relati extrema. To produce a plot that reveals the true structure of the graph, we replot over the interval [-1,2]: >> x=linspace (-1,2); >> y= 4 * x. ^4-12*x.^3 >> plot (x, y), grid 91x."2; + We see two relative minima, one relative maximum and two points of inflection. Next, we plot the derivative function, f(x), which we name yp for short, on the same graph as f(x) in red > hold on 16"x.^3-36*x."2+18*x; >> yp s> plot (x, yp, 'r') We know that the zoom method works poorly to find the exact value of a relative maximum or minimum, as it is hard to find out exactly where the true graph is flat from the approximation that MATLAB produces. However, at a relative extrema, the function has a critical point, so in many cases the derivative is zero. We can then zoom in on the graph of f (x) to investigate the relative extrema. Exercise 1: 1. Find the zeros of f(x) correct to two decimal places. correct ou received (0,0.75,1.5) 100% of 10 points

Answer 1

The zeros we found earlier are 0, 0.75 and 1.5

To find the co-ordinates of the point corresponding to these points, we have:

x-0, 0.75, 1.5); % the y=41x.^4-121x.^3491x."2 points %value >> of function on these point s >> 0 1.2656

$(x,y)=(0.75,1.2656)$ is the relative maxima's co-ordinates