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3. mydetecttop(f,a,t) which takes three inputs: [50 pts] f: A function handle. a: A real number. t: A real number. Note: You may assume that f is increasing at x = a but there is a cutoff x = b > a after which f starts decreasing. Does: Uses a while loop to check f(x) for x = a, x = a + 3t, x = a + 6t, ... until a value is smaller than the previous value. Returns: The x value (not the f(x) value!) for which this happens

Respuesta :

ammary456

Answer:

function n = mydetecttop(f, a, t)

n = a;

 

value1 = f(n);

i1 = 4;

 

while(true)

 

n = a + i1*t;

value2 = f(n);

 

if(value2 < value1)

   

break

end

 

value1 = value2;

 

i1 = i1 + 4;

end

 

end

function t = atc(n)

 

if(n<25)

t = n + 1;

else

t = 0 - n;

end

end

f = @atc;

n = mydetecttop(f, 1, 2)

Explanation:

  • Run a while loop until the  required value is found and update the value of  the variable n.
  • Check if value2 is less than value1 and then break out of the loop.
  • Update the value of  the variable i1 by adding 4.
  • Finally test the mydetecttop(f,a,t) function.

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