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Why in rolles theorem closed interval is used for contuinity and open interval is used for differentiability?
It requires that f(a)=f(b) where a and b are beginning and ending points. Also, it says there is a c between a and such that f'(c)=0. If f were not differentiable on the open interval, the statement f'(c)=0 would be invalid.
What is the difference between Taylor series and Maclaurin series?
A Maclaurin series is centered about zero, while a Taylor series is centered about any point c. M(x) = [f(0)/0!] + [f'(0)/1!]x +[f''(0)/2!](x^2) + [f'''(0)/3!](x^3) + . . . for f(x). T(x) = [f(c)/0!] + [f'(c)/1!](x-c) +[f''(c)/2!]((x-c)^2) + [f'''(c)/3!]((x-c)^3) + . . . for f(x).
Solve ex-3x2 equals 0 by bisection method?
I'm assuming you're trying to write f(x)=exp(x)-3x2=0 as your function. This has 3 solutions where f(x)=0 between (-0.5,-0.4), (0.85, 0.95) and (3.7, 3.8) I won't find all of the roots, only the one between 3.7 and 3.8 in order to show you the procedure/algorithm.What we do is choose two points x=a, b on the function where it takes on opposite values. We then find the mid point (a+b)/2 between these values and find the value of the functions at f((a+b)/2) we then choose the smallest interval where the points take on opposite values and repeat until we have a small enough value for x.We will do our iteration until we get within 0.01 of the solution (i.e. the first number to give an answer between -0.01 and 0.01) and i will be using c instead of writing out (a+b)/2 all the time.iteration 1: a=3.7 b=3.8 c=3.75 f(c)=0.3335 +veiteration 2: a=3.7 b=3.75 c=3.725 f(c)=-0.155 -veiteration 3: a=3.725 b=3.75 c=3.7375 f(c)=0.086 +veiteration 4: a=3.725 b=3.7375 c=3.73125 f(c)=-0.035 -veiteration 5: a=3.73125 b=3.7375 c=3.734375 f(c)=0.025 +veiteration 6: a=3.73125 b=3.734375 c=3.732813 f(c)=-0.005 between -0.01 and 0.01ONE solution for f(x)=exp(x)-3x2=0 is x=3.732813 (actual value is x=3.733079029 so we have an absolute error of 0.000266029, a relative error of 0.00007126262207 and a percentage error of 0.007126262207%)
How do you convert the temperature between Celsius and Fahrenheit?
C = ( F - 32 ) x 5/9orF = C x 9/5 +32Where F is degrees Fahrenheit and C degrees Celsius
A quantity that depends on force distance and time is called?
That's power.P = FS (theta)/T; where F is force, S is distance, T is time, and theta is the angle between F and S.
