Find the absolute minimum value for on the interval [-1,1] f(x)=x^{3}-6x^{2}-15x+3

Anne Wacker

Anne Wacker

Answered question

2021-12-12

Find the absolute minimum value for on the interval [-1,1]
f(x)=x36x215x+3

Answer & Explanation

Debbie Moore

Debbie Moore

Beginner2021-12-13Added 43 answers

Step 1
In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value.
The most basic and common algebraic equations in math consist of one or more variables.
For instance, 4x+10=7 is an equation, in which 4x+10 and 7 are two expressions separated by an ‘equal’ sign.
Step 2
Differentiate the given equation and then equate it to 0 to obtain the extreme values.
f(x)=x36x215x+3
f(x)=d(x36x215x+3)dx
=3x212x15
3x212x15=0
x24x5=0
Step 3
Now solve the equation x24x5=0 using the factorization method and obtain the value of x.
x24x5=0
x2+x5x5=0
x(x+1)-5(x=1)=0
(x+1)(x-5)=0
x=-1
or
x=5
Therefore the absolute minimum value is -1

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