Suppose F is a differentiable function such that f'(x)≤=5 for all ...?
Suppose F is a differentiable function such that f'(x)≤=5 for all x E [-1,4]. If f(-1)=2 , the Mean Value Theorem says that f(4)≤=V for what value of V? (Choose V as small as possible.)
the diversity of f(x) = x^5+x^3+x^2+x is y?R. f additionally has only one answer, x=0. The max and min of cos(x) is a million and -a million. So while cos(x) is extra to f, any zeroes of the ensuing function, g(x) = x^5+x^3+x^2+x+cos(x) might could take place interior of a internet site such that -a million < f < a million f(0.5295) = a million and f(-0.8) = -a million 0< f < a million while x?(0, 0.5295) and a million < cos(x) < 0.8630597 because f is effective in this region, purely detrimental values of cos(x) can pull f all the way down to furnish a nil, yet because cos(x) isn't detrimental, no suggestions are achieveable staring at in simple terms the detrimental x-values because thats of difficulty in this issue and the effective x-values do no longer furnish any suggestions, => -a million < f < 0 while x?(-0.8, 0) additionally 0.696707 < cos(x) < a million while x?(-0.8, 0) so what we'd desire to look for in this area are factors the place f(x) = -cos(x) the only different region the place a nil ought to take place replaced into x?(0, 0.5295), even though it replaced into proved that no zeroes befell right here. simply by fact the sum of wierd applications is likewise ordinary, the function g additionally must be ordinary and subsequently could have a minimum of one 0. because x?(-0.8, 0) is the only different region it could have a nil, the 0 ought to be detrimental. Say one 0 of the function g happens at x=c (-0.8<c<0), at this element f = -cos(x). If this variety of concern (f = -cos(x)), has to take place lower back, f and cos(x) might could enhance in opposite instructions or no longer enhance in any respect (i.e. proceed to be consistent), the two for x<c or for x>c. yet from the by-made from the two f and cos(x), all of us understand that they the two boost in x?(-0.8, 0). subsequently there can purely be one root. consequently we've proved that g has purely one detrimental root. Q.E.D playstation . i've got not been very concise, its in simple terms that I certainly have never accomplished a information like this in the past, so i used as many words as i presumed replaced into had to do the information. additionally i've got used approx values, be happy to alter them with right values. It should not be too complicated to video show a similar consequences while employing right values because cos(x) is a "incredibly" function and all of us understand all its maxes, minutes and roots. Cheers!
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Verified answer
slope is at most 5
then lets suppose it IS 5 all across the interval
then we have a line from the point (-1,2)
2= 5 (-1) +c
c=2+5=7
y=5x+7
now for x=4
y=20+7=27
V=27
the diversity of f(x) = x^5+x^3+x^2+x is y?R. f additionally has only one answer, x=0. The max and min of cos(x) is a million and -a million. So while cos(x) is extra to f, any zeroes of the ensuing function, g(x) = x^5+x^3+x^2+x+cos(x) might could take place interior of a internet site such that -a million < f < a million f(0.5295) = a million and f(-0.8) = -a million 0< f < a million while x?(0, 0.5295) and a million < cos(x) < 0.8630597 because f is effective in this region, purely detrimental values of cos(x) can pull f all the way down to furnish a nil, yet because cos(x) isn't detrimental, no suggestions are achieveable staring at in simple terms the detrimental x-values because thats of difficulty in this issue and the effective x-values do no longer furnish any suggestions, => -a million < f < 0 while x?(-0.8, 0) additionally 0.696707 < cos(x) < a million while x?(-0.8, 0) so what we'd desire to look for in this area are factors the place f(x) = -cos(x) the only different region the place a nil ought to take place replaced into x?(0, 0.5295), even though it replaced into proved that no zeroes befell right here. simply by fact the sum of wierd applications is likewise ordinary, the function g additionally must be ordinary and subsequently could have a minimum of one 0. because x?(-0.8, 0) is the only different region it could have a nil, the 0 ought to be detrimental. Say one 0 of the function g happens at x=c (-0.8<c<0), at this element f = -cos(x). If this variety of concern (f = -cos(x)), has to take place lower back, f and cos(x) might could enhance in opposite instructions or no longer enhance in any respect (i.e. proceed to be consistent), the two for x<c or for x>c. yet from the by-made from the two f and cos(x), all of us understand that they the two boost in x?(-0.8, 0). subsequently there can purely be one root. consequently we've proved that g has purely one detrimental root. Q.E.D playstation . i've got not been very concise, its in simple terms that I certainly have never accomplished a information like this in the past, so i used as many words as i presumed replaced into had to do the information. additionally i've got used approx values, be happy to alter them with right values. It should not be too complicated to video show a similar consequences while employing right values because cos(x) is a "incredibly" function and all of us understand all its maxes, minutes and roots. Cheers!