What are the extrema of #f(x)=-3x^2+30x-74# on #[-oo,oo]#?

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Answer 1 · 2017-10-11T07:03:55

Let's see.

Let the function given be #y# such that #rarr# for any value of #x# in the given range.

#y=f(x)=-3x^2+30x-74#

#:.dy/dx=-6x+30#

#:.(d^2y)/dx^2=-6#

Now, since the second order derivative of the function is negative, the value of #f(x)# will be maximum.

Hence, point of maxima or extrema can only be obtained.

Now, whether for maxima or minima,

#dy/dx=0#

#:.-6x+30=0#

#:.6x=30#

#:.x=5#

Therefore, the point of maxima is #5#. (Answer).

So, the maximum value or the extreme value of #f(x)# is #f(5)#.

#:.f(5)=-3.(5)^2+30.5-74#

#:.f(5)=-75+150-74#

#:.f(5)=150-149#

#:.f(5)=1#.

Hope it Helps:)