Consider these methods: public static double f(double x) { return g(x) + Math.sq
ID: 3530188 • Letter: C
Question
Consider these methods: public static double f(double x) { return g(x) + Math.sqrt(h(x)); } public static double g(double x) { return 4 * h(x); } public static double h(double x) { return x * x + k(x) - 1; } public static double k(double x) { return 2 * (x + 1); } Without actually compiling and running a program, determine the results of the following method calls: double x1 = f(2); double x2 = g(h(2)); double x3 = k(g(2) + h(2)); double x4 = f(0) + f(1) + f(2); double x5 = f(-1) + g(-1) + h(-1) + k(-1);Explanation / Answer
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1) double x1 = f(2) = g(2)+sqrt(h(2)) = 4*h(2)+sqrt(h(2))
h(2) = 4+k(2)-1 = 3+2*3 =9;
so x1= 4*9+sqrt(9) =39
2) double x2 = g(h(2)) = g(9) // h(2)=9 from above result
=4*h(9) = 4*(9*9+k(9)-1) = 4*(80+20) =400
so x2=400
3) double x3= k(g(2)+h(2)) = 2*(g(2)+h(2)+1)
= 2*(4*h(2)+h(2)+1) =2*(5*h(2)+1)
= 2*(45+1)//h(2)=9
= 92
so x3=92
4) double x4= f(0)+f(1)+f(2) = 5+18+39 = 72
5) double x5 = f(-1)+g(-1)+h(-1)+k(-1) = 0+0+0+0 =0