Question

The maximum value of the function y=4x2+3x+3 where ‘x ’ varies from ‘0’ to ‘4’:-

A

118

B

79

C

3

D

10

​

Answer 1

Answer:

b=79

Step-by-step explanation:

as we know that for finding the maximum value in the given interval we need to find the derivative for the given function.

$y = 4 {x}^{2} + 3x + 3$

now derivative is

$8x + 3 = 0$

$x = - \frac{3}{8}$

so the given function as the extrima at x=0,4,

$- \frac{3}{8}$

hence now check the function at all three points and we get the highest value of the function is 79.

Answer 2

Step-by-step explanation:

try to solve problems with this solution