If f(x) is a differentiable function such that f '(0) = 2, f '(2) = -3 and f '(5) = 7 then the limit lim ( [f(x) - f(4)] / (x - 4) ) as x approaches 4 ?
if we put 4 in place of x the we get 0/0 form so we need to differentiate numerator and denominator separately. => lim ( [f(x) - f(4)] / (x - 4) ) = lim ( [f ' (x) / 1) as f(x) - f(4) when differentiated we get f ' (x) as differentiation of constant is zero. Similarly differentiation of x - 4 is 1 as differentiation of x is 1 and constant is 0. lim ( [f(x) - f(4)] / (x - 4) ) as x approaches 4 = lim ( f ' (x) ) as x approaching 4 the answer is f ' (4) which cannot be found from the given information. so more information is needed.
What are the different ways of passing parameters to the functions? Which to use when?
--Call by value We send only values to the function as parameters. We choose this if we do not want the actual parameters to be modified with formal parameters but just used. --Call by reference We send address of the actual parameters instead of values. We choose this if we do want the actual parameters to be modified with formal parameters.
If f(x) = 5 - (2^x), let g(x) be the inverse of function f(X). then g(-3) = ? ( '^' indicates power function)
First we need to find g(x) from given f(x) and using inverse relation. y = 5 - 2^x , ****given and let f(x) be denoted by 'y'**** x = 5 - 2^y , ****interchange x and y in order to inverse of f(x) because inverse mean domain become co-domain and co-domain is domain.**** Solving for y now 2^y = 5 - x , ****apply log to the base two on both sides**** y = log(5-x), ***** here log base is 2***** now in place of y we can replace inverse of f(x) which is g(x) => g(x) = log (5 - x) ****(BASE 2)**** g(-3) = log( 5 - (-3) ) g(-3) = log(8) = log ( (2) ^ 3) ) *** 2 cube is 2^3 = 8***** g(-3) = 3 *** log (x^a) = a when base is x for the log***