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

Question

Christopher Allers · Accepted Answer

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:)

Ezekiel Alexander · Answer

undefined To find the extrema of $ f(x) = -3x^2 + 30x - 74 $ on the interval $[- \infty, \infty]$, we first find the critical points by taking the derivative and setting it to zero:

$$ f'(x) = -6x + 30 $$

Setting $ f'(x) $ to zero gives:

$$ -6x + 30 = 0 $$

$$ -6x = -30 $$

$$ x = 5 $$

So, $ x = 5 $ is a critical point. To determine if it is a local maximum or minimum, we can use the second derivative test.

$$ f''(x) = -6 $$

Since $ f''(x) $ is negative for all $ x $, the critical point $ x = 5 $ corresponds to a local maximum. Since there are no other critical points and the function is a downward facing parabola, this local maximum is also the global maximum on the interval $[- \infty, \infty]$.