`f(x)=4x^3+19x^2-14x+3`
Find Local maxima and minima

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)`

`:.` The critical points are `x=(1/3)` and `x=-(7/2)`

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Step-3: Apply the second derivative test
Now, `f^('')(x)=``d/(dx)(12x^2+38x-14)`

`=d/(dx)(12x^2)+d/(dx)(38x)-d/(dx)(14)`

`=24x+38-0`

`=24x+38`

Evaluate `f^('')(x)` at the critical points

For `x=(1/3)`

`f^('')((1/3))``=24*(1/3)+38`

`=8+38`

`=46`` > 0`

`:.` At `x=(1/3)` the function is local minimum

For `x=-(7/2)`

`f^('')(-(7/2))``=24*(-(7/2))+38`

`=24*(-3.5)+38`

`=-84+38`

`=-46`` < 0`

`:.` At `x=-(7/2)` the function is local maximum

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Step-4: Calculate the extrema values
Substitute the `x` values back into the original function `f(x)`

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

1. At `x=(1/3)`

`f((1/3))``=4*(1/3)^3+19*(1/3)^2-14*(1/3)+3`

`=0.1481+2.1111-4.6667+3`

`=0.5926`

local minimum point = `((1/3),0.5926)`

2. At `x=-(7/2)`

`f(-(7/2))``=4*(-(7/2))^3+19*(-(7/2))^2-14*(-(7/2))+3`

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

`=-171.5+232.75+49+3`

`=113.25`

local maximum point = `(-(7/2),113.25)`

graph

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