find the local maxima and minima for the function y=x³-3x+2
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Answer:
Given, Function,y=x3−3x2+6
So for that, we have to find the critical value of y so, differentiate y respect to x
y′=3x2−6x=3x(x−2)=0
So, X is either be 2 or 0. these are the critical values.
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Calculus
Find the Local Maxima and Minima y=x^3-3x+2
y=x3−3x+2y=x3-3x+2
Rewrite the equation as a function of xx.
f(x)=x3−3x+2f(x)=x3-3x+2
Find the first derivative of the function.
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3x2−33x2-3
Find the second derivative of the function.
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f''(x)=6xf′′(x)=6x
To find the local maximum and minimum values of the function, set the derivative equal to 00 and solve.
3x2−3=03x2-3=0
Add 33 to both sides of the equation.
3x2=33x2=3
Divide each term by 33 and simplify.
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x2=1x2=1
Take the square root of both sides of the equation to eliminate the exponent on the left side.
x=±√1x=±1
The complete solution is the result of both the positive and negative portions of the solution.
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x=1,−1x=1,-1
Evaluate the second derivative at x=1x=1. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(1)6(1)
Multiply 66 by 11.
66
x=1x=1 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=1x=1 is a local minimum
Find the y-value when x=1x=1.
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y=0y=0
Evaluate the second derivative at x=−1x=-1. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
6(−1)6(-1)
Multiply 66 by −1-1.
−6-6
x=−1