Locate relative maxima and minima f(x)=-3/4x^4 +1/3x^3 +5x^2 +3

Question

Supposed to use the first derivative test.Can you please show me the steps. Thanks!

Answer 1

The function is f ( x ) = ( - 3 / 4 ) x ⁴ + ( 1 / 3 ) x ³ + 5 x ²+ 3

Derivative both sides with respected to x .

f ' ( x ) = d / dx [ ( - 3 / 4 ) x ⁴ + ( 1 / 3 ) x ³ + 5 x ²+ 3 ]

The power prorerty  in derivatives d / dx  x ⁿ= n * x ^{n - 1}

f ' ( x ) = d / dx ( - 3 / 4 ) x ⁴ + d / dx ( 1 / 3 ) x ³+d / dx ( 5 x ² ) + d / dx ( 3 ) 

=( - 3 / 4 ) d / dx ( x ⁴ )+ ( 1 / 3 ) d / dx(  x ³) + 5 d / dx(  x ²) + 0

= ( - 3 / 4 ) (4 x ^{4 - 1} ) + ( 1 / 3 ) (3 x ^{3 - 1} ) + 5 * 2 x

= ( - 3 / 4 ) (4 x ³ ) + ( 1 / 3 ) ( 3 x ²  ) + 10 x

= - 3 x ³ +  x ²  + 10 x 

= x  (- 3 x ²+  x + 10)

= x  (- 3 x ²+ 6 x  + 5 x + 10)

= x [ (- 3 x ( x -2 ) - 5( x - 2) ]

= x [ ( x - 2 )(- 3x - 5) ]

f ' ( x ) = x [ ( x - 2 )(- 3x - 5) ].

To find the critical points, to make the first derivative equal to zero or f ' (x) = 0.

x [ ( x - 2 )(- 3x - 5) ] = 0.

x = 0 , ( x - 2 ) = 0 , (- 3x - 5) = 0

x = 0 , x  = 2 , x = - 5 / 3

 x = 0 : -

f (0 ) = 3

if x = 2 then f ( 2 ) = ( - 3 / 4 ) 2 ⁴ + ( 1 / 3 ) 2 ³ + 5 2 ²+ 3

= ( - 3 / 4 ) 16 + ( 1 / 3 ) 8 + 5 * 4 + 3

= -12 + ( 8 / 3 ) + 20 + 3

=(  - 36 +  8  + 60 + 9 ) / 3

=(  - 36 + 77 ) / 3

=( 41 ) / 3 = 13 . 66

x = - 5 / 3 : -

f ( 2 ) = ( - 3 / 4 ) ( - 5 / 3 ) ⁴ + ( 1 / 3 ) (- 5 / 3) ³ + 5 ( - 5 / 3 ) ²+ 3

= ( - 3 / 4 )( 625 / 81 )+ ( 1 / 3 ) (- 125 / 27) ³ + 5 ( 25 / 9 ) + 3

= 9 . 55

Max { f ( x ) = ( - 3 / 4 ) x ⁴ + ( 1 / 3 ) x ³ + 5 x ²+ 3 } = 41 / 3 at  x = 2.

Min { f ( x ) = ( - 3 / 4 ) x ⁴ + ( 1 / 3 ) x ³ + 5 x ²+ 3 } =  3 at  x = 0.

Comments

sorry !___ typo mistake  -------->

f ( - 5/3 ) = ( - 3 / 4 ) ( - 5 / 3 ) ⁴ + ( 1 / 3 ) (- 5 / 3) ³ + 5 ( - 5 / 3 ) ²+ 3

= ( - 3 / 4 )( 625 / 81 )+ ( 1 / 3 ) (- 125 / 27) ³ + 5 ( 25 / 9 ) + 3

= 9 . 55

Max { f ( x ) = ( - 3 / 4 ) x ⁴ + ( 1 / 3 ) x ³ + 5 x ²+ 3 } = 41 / 3 at  x = 2.

Min { f ( x ) = ( - 3 / 4 ) x ⁴ + ( 1 / 3 ) x ³ + 5 x ²+ 3 } =  3 at  x = 0.