`f(x)=4x^3+19x^2-14x+3`
Find Local maxima and minima of a function using first derivative test

Solution:
Here, `f(x)=4x^3+19x^2-14x+3`

Step-1: Find the derivative of the function
`:. f^'(x)=``d/(dx)(4x^3+19x^2-14x+3)`

`=d/(dx)(4x^3)+d/(dx)(19x^2)-d/(dx)(14x)+d/(dx)(3)`

`=12x^2+38x-14+0`

`=12x^2+38x-14`

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Step-2: Find the critical points of the derivative function
To find critical points, set `f^'(x)=0` and then solve for x

`f^'(x)=0`

`=>12x^2+38x-14 = 0`

`=>2(6x^2+19x-7) = 0`

`=>2(6x^2-2x+21x-7) = 0`

`=>2(2x(3x-1)+7(3x-1)) = 0`

`=>2(3x-1)(2x+7) = 0`

`=>(3x-1) = 0" or "(2x+7) = 0`

`=>3x = 1" or "2x = -7`

`=>x = 1/3" or "x = -7/2`

The solution is
`x = (1/3),x = -(7/2)`

`:.` `x=-3.5` and `x=0.3333`

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Step-3: Use the critical points to determine intervals
There are total 2 critical points, So we have 3 intervals
`(-oo,-3.5),(-3.5,0.3333),(0.3333,oo)`

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Step-4: Determine the Sign of `f^'(x)` in each interval

1. For first interval `(-oo,-3.5)`, we choose `x=-4.5`

`f^'(-4.5)``=12*(-4.5)^2+38*(-4.5)-14`

`=243-171-14`

`=58`` > 0`

2. For second interval `(-3.5,0.3333)`, we choose `x=0`

`f^'(0)``=12*0^2+38*0-14`

`=0+0-14`

`=-14`` < 0`

3. For third interval `(0.3333,oo)`, we choose `x=1.3333`

`f^'(1.3333)``=12*1.3333^2+38*1.3333-14`

`=21.3323+50.6654-14`

`=57.9977`` > 0`

Interval	x-value	`f^'(x)`	Negative or Positive
`(-oo,-3.5)`	`-4.5`	`f^'(-4.5)=58`` > 0`	Positive
`(-3.5,0.3333)`	`0`	`f^'(0)=-14`` < 0`	Negative
`(0.3333,oo)`	`1.3333`	`f^'(1.3333)=57.9977`` > 0`	Positive

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Step-5: Conclude the nature of each critical point
At `x=-3.5`, `f^'(x)` changes from positive to negative, indicating a local maximum

At `x=0.3333`, `f^'(x)` changes from positive to negative, indicating a local minimum

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Step-6: Determine the function values at the critical points
Calculate `f(x)` at the critical points to find the maximum and minimum values

1. At `x=-3.5`

`f(-3.5)``=4*(-3.5)^3+19*(-3.5)^2-14*(-3.5)+3`

`=-171.5+232.75+49+3`

`=113.25`

Local Maximum point = `(-3.5,113.25)`

2. At `x=0.3333`

`f(0.3333)``=4*0.3333^3+19*0.3333^2-14*0.3333+3`

`=0.1481+2.1107-4.6662+3`

`=0.5926`

Local Minimum point = `(0.3333,0.5926)`

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