Question

for what value of X does the function f(X)=X3-9x2 -120 X +6 have a local minimum?​

Answer 1

Given

f(x)= x³-9x²-120x+6

To find

We have to find local minimum

$\sf\huge {\underline{\underline{{Solution}}}}$

$\sf\implies \: f(x) = {x}^{3} - 9 {x}^{2} - 120x + 6$

Now, we find derivatives of the given function:

$\sf\implies \: {f}^{'}( x) = 3 \times {x}^{2} - 2 \times 9x - 120(1)$

Identity used here :

$\sf\boxed{ {x}^{n} = {nx}^{n - 1} }$

Note : The derivative of any constant is zero.

=> f'(x) = 3x²-18x-120

Now,we factorise it

=> 3(x²-6x-40)=0

=> x²-6x-40

=> x²-10x+4x-40

=>x(x-10)+4(x-10)=0

=> (x+4)(x-10)=0

x= -4 & x= 10

Now, finding second order derivatives:

=>f"(x)= 3x²-18x-120

=>f"(x)= 6x-18

f"(x)=6x-18

Now,put the value of x into f"(x)

At x = -4

=>f"(x)= 6(-4)-18

=>f"(x)= -24-18

=>f"(x)= -42

f"(x) is < 0

At x= 10

=>f"(x)= 6(10)-18

=>f"(x)= 60-18= 42

f"(x) is > 0

Concept:

• When the f"(x) is less than zero then it will be maxima and the points we obtained is called its local maximum points.
• when the f"(x) is greater than zero then it will be minima and the points we obtained is called its local minimum Points.

Since , f"(x) = 42 which is greater than zero.

Therefore,the local minimum Point is 42

Answer 2

Answer:

Step-by-step explanation:

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