Question

absolute maximum of the function 3x + 5 in (6,12) is​

Answer 1

Answer:

Step-by-step explanation:

f(x)=3x  

4

−8x  

3

+12x  

2

−48x+25[0,3]

f  

′

(x)=12x  

3

−24x  

2

+24x−48

For critical points f  

′

(x)=0

⇒12x  

3

−24x  

2

+24x−48=0

⇒12(x  

3

−2x  

2

+2x−4)=0

⇒12(x  

2

+2)(x−2)=0

⇒x=2

f(0)=25

f(2)=48−64+48−96+25=−39

f(3)=243−216+108−144+25=16

Absolute maximum =25 at x=0

Absolute minimum =−39 at x=2.

Answer 2

Given,

$\[f\left( x \right) = 3x + 5\]$

Solution,

$\[\begin{array}{l}f\left( x \right) = 3x + 5\\ \Rightarrow f'\left( x \right) = 3\end{array}\]$

$\[ \Rightarrow f'\left( x \right) = + ve\]$

Calculate the absolute maximum value of the function.

$\[\begin{array}{l}f\left( x \right) = 3x + 5\\ \Rightarrow f\left( {x = 6} \right) = 3 \times 6 + 5\\ \Rightarrow f\left( {x = 6} \right) = 23\\ \Rightarrow f\left( {x = 12} \right) = 3 \times 12 + 5\\ \Rightarrow f\left( {x = 12} \right) = 41\end{array}\]$

Hence the absolute maximum value of function is $\[41\]$.