For f(x)= 3x^3 + 4 x^2 + 3x +10 , if f’(x) =0 what can be the values

of x and f(x)?

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To solve this question we first find the derivative of f(x). Here we use the relation that for f(x)= a x^n,

f(x) = a*n*x^(n-1)

As f(x) = 3x^3 + 4 x^2 + 3x +10

f’(x) = 9x^2 + 8x + 3

Now if f’(x)= 0

=> 9x^2 + 8x + 3 =0

=> x = [-b+sqrt( b^2 – 4ac)]/2a and x= [-b-sqrt( b^2 – 4ac)]/2a

=> x = [-8 + sqrt( 64 -108)]/18 and x= [-8 - sqrt( 64 -108)]/18

=> x = -4/9 + (sqrt 44*i) / 18 and x= -4/9 - (sqrt 44*i) / 18

For x =-4/9 + (sqrt 44*i)/18, f(x) = 2234/243 - 0.3i

For x =-4/9 - (sqrt 44*i)/18, f(x)= 2234/243 + 0.3i

f(x) = 3x^3+4x^2+3x+10 and f'(x) = 0.

To find values of x and f(x).

Solution:

So we equate the differential coefficient of x to zero and solve the resulting (quadratic) equation for x . And then find the f(x).

So (3x^3+4x^2+3x+10)' = 0

3*3x^2+4*2x+3 = 0

9x^2+8x+3 = 0

But the discriminant of the equation is 8^2- 4*9*3 = 64-108 = -44 < 0.

Therefore x is imaginary. So f(x) is also imaginary or does not exist in real value .

f'(x)=3*3x^2+4*2x+3*1x^0+0

f'(x)=9x^2+8x+3

f'(x)=0 9x^2+8x+3=0

x=(-b-(b^2-4ac)^1/2)/2a=(-8-(8^2-4*9*3)^1/2)/2*9= imposible

x=(-b+(b^2-4ac)^1/20/2a=(-8+(8^2-4*9*3)^1/2)/2*9= imposible

X is not exist

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