Question

for any real number x, the maximum value of 4- 6x-x^2 is —​

Answer 1

Answer:

13

Step-by-step explanation:

f(x)=4−6x−x

2

For critical points,

f ′ (x)=0

−6−2x=0

−2x=6

x=−3

For maxima/minima

f

′′

(x)=−2 (negative)

Hence at x=−3f(x) will have maxima

f(−3)=4−6×(−3)−(−3)

2

=4+18−9

f(−3)=13

Maximum value of f(x) is 13