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

1 Answer
Oct 11, 2017

Answer:

Let's see.

Explanation:

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:)