Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=-3|x|
f(x)=-3|x|
Step 1
Find the first derivative of the function.
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Step 1.1
Since -3 is constant with respect to x, the derivative of -3|x| with respect to x is -3ddx[|x|].
-3ddx[|x|]
Step 1.2
The derivative of |x| with respect to x is x|x|.
-3x|x|
Step 1.3
Combine fractions.
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Step 1.3.1
Combine -3 and x|x|.
-3x|x|
Step 1.3.2
Move the negative in front of the fraction.
-3x|x|
-3x|x|
-3x|x|
Step 2
Find the second derivative of the function.
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Step 2.1
Since -3 is constant with respect to x, the derivative of -3x|x| with respect to x is -3ddx[x|x|].
f′′(x)=-3ddxx|x|
Step 2.2
Differentiate using the Quotient Rule which states that ddx[f(x)g(x)] is g(x)ddx[f(x)]-f(x)ddx[g(x)]g(x)2 where f(x)=x and g(x)=|x|.
f′′(x)=-3|x|ddx(x)-xddx|x||x|2
Step 2.3
Differentiate using the Power Rule.
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Step 2.3.1
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
f′′(x)=-3|x|1-xddx|x||x|2
Step 2.3.2
Multiply |x| by 1.
f′′(x)=-3|x|-xddx|x||x|2
f′′(x)=-3|x|-xddx|x||x|2
Step 2.4
The derivative of |x| with respect to x is x|x|.
f′′(x)=-3|x|-xx|x||x|2
Step 2.5
Combine x|x| and x.
f′′(x)=-3|x|-xx|x||x|2
Step 2.6
Raise x to the power of 1.
f′′(x)=-3|x|-xx|x||x|2
Step 2.7
Raise x to the power of 1.
f′′(x)=-3|x|-xx|x||x|2
Step 2.8
Use the power rule aman=am+n to combine exponents.
f′′(x)=-3|x|-x1+1|x||x|2
Step 2.9
Add 1 and 1.
f′′(x)=-3|x|-x2|x||x|2
Step 2.10
Combine -3 and |x|-x2|x||x|2.
f′′(x)=-3(|x|-x2|x|)|x|2
Step 2.11
Move the negative in front of the fraction.
f′′(x)=-3(|x|-x2|x|)|x|2
Step 2.12
Simplify.
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Step 2.12.1
Apply the distributive property.
f′′(x)=-3|x|+3(-x2|x|)|x|2
Step 2.12.2
Simplify each term.
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Step 2.12.2.1
Multiply 3(-x2|x|).
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Step 2.12.2.1.1
Multiply -1 by 3.
f′′(x)=-3|x|-3x2|x||x|2
Step 2.12.2.1.2
Combine -3 and x2|x|.
f′′(x)=-3|x|+-3x2|x||x|2
f′′(x)=-3|x|+-3x2|x||x|2
Step 2.12.2.2
Move the negative in front of the fraction.
f′′(x)=-3|x|-3x2|x||x|2
f′′(x)=-3|x|-3x2|x||x|2
Step 2.12.3
Simplify the numerator.
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Step 2.12.3.1
Factor 3 out of 3|x|-3x2|x|.
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Step 2.12.3.1.1
Factor 3 out of 3|x|.
f′′(x)=-3(|x|)-3x2|x||x|2
Step 2.12.3.1.2
Factor 3 out of -3x2|x|.
f′′(x)=-3(|x|)+3(-x2|x|)|x|2
Step 2.12.3.1.3
Factor 3 out of 3(|x|)+3(-x2|x|).
f′′(x)=-3(|x|-x2|x|)|x|2
f′′(x)=-3(|x|-x2|x|)|x|2
Step 2.12.3.2
To write |x| as a fraction with a common denominator, multiply by |x||x|.
f′′(x)=-3(|x||x||x|-x2|x|)|x|2
Step 2.12.3.3
Combine the numerators over the common denominator.
f′′(x)=-3(|x||x|-x2|x|)|x|2
Step 2.12.3.4
Simplify the numerator.
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Step 2.12.3.4.1
Multiply |x||x|.
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Step 2.12.3.4.1.1
To multiply absolute values, multiply the terms inside each absolute value.
f′′(x)=-3(|xx|-x2|x|)|x|2
Step 2.12.3.4.1.2
Raise x to the power of 1.
f′′(x)=-3(|xx|-x2|x|)|x|2
Step 2.12.3.4.1.3
Raise x to the power of 1.
f′′(x)=-3(|xx|-x2|x|)|x|2
Step 2.12.3.4.1.4
Use the power rule aman=am+n to combine exponents.
f′′(x)=-3(|x1+1|-x2|x|)|x|2
Step 2.12.3.4.1.5
Add 1 and 1.
f′′(x)=-3(|x2|-x2|x|)|x|2
f′′(x)=-3(|x2|-x2|x|)|x|2
Step 2.12.3.4.2
Remove non-negative terms from the absolute value.
f′′(x)=-3(x2-x2|x|)|x|2
Step 2.12.3.4.3
Subtract x2 from x2.
f′′(x)=-3(0|x|)|x|2
f′′(x)=-3(0|x|)|x|2
Step 2.12.3.5
Divide 0 by |x|.
f′′(x)=-30|x|2
f′′(x)=-30|x|2
Step 2.12.4
Remove the absolute value in |x|2 because exponentiations with even powers are always positive.
f′′(x)=-30x2
Step 2.12.5
Multiply 3 by 0.
f′′(x)=-0x2
Step 2.12.6
Divide 0 by x2.
f′′(x)=-0
Step 2.12.7
Multiply -1 by 0.
f′′(x)=0
f′′(x)=0
f′′(x)=0
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
-3x|x|=0
Step 4
Find the first derivative.
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Step 4.1
Find the first derivative.
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Step 4.1.1
Since -3 is constant with respect to x, the derivative of -3|x| with respect to x is -3ddx[|x|].
-3ddx[|x|]
Step 4.1.2
The derivative of |x| with respect to x is x|x|.
-3x|x|
Step 4.1.3
Combine fractions.
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Step 4.1.3.1
Combine -3 and x|x|.
-3x|x|
Step 4.1.3.2
Move the negative in front of the fraction.
f(x)=-3x|x|
f(x)=-3x|x|
f(x)=-3x|x|
Step 4.2
The first derivative of f(x) with respect to x is -3x|x|.
-3x|x|
-3x|x|
Step 5
Set the first derivative equal to 0 then solve the equation -3x|x|=0.
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Step 5.1
Set the first derivative equal to 0.
-3x|x|=0
Step 5.2
Set the numerator equal to zero.
3x=0
Step 5.3
Divide each term in 3x=0 by 3 and simplify.
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Step 5.3.1
Divide each term in 3x=0 by 3.
3x3=03
Step 5.3.2
Simplify the left side.
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Step 5.3.2.1
Cancel the common factor of 3.
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Step 5.3.2.1.1
Cancel the common factor.
3x3=03
Step 5.3.2.1.2
Divide x by 1.
x=03
x=03
x=03
Step 5.3.3
Simplify the right side.
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Step 5.3.3.1
Divide 0 by 3.
x=0
x=0
x=0
Step 5.4
Exclude the solutions that do not make -3x|x|=0 true.
No solution
No solution
Step 6
Find the values where the derivative is undefined.
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Step 6.1
Set the denominator in 3x|x| equal to 0 to find where the expression is undefined.
|x|=0
Step 6.2
Solve for x.
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Step 6.2.1
Remove the absolute value term. This creates a ± on the right side of the equation because |x|=±x.
x=±0
Step 6.2.2
Plus or minus 0 is 0.
x=0
x=0
x=0
Step 7
Critical points to evaluate.
x=0
Step 8
Evaluate the second derivative at x=0. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
0
Step 9
Since there is at least one point with 0 or undefined second derivative, apply the first derivative test.
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Step 9.1
Split (-,) into separate intervals around the x values that make the first derivative 0 or undefined.
(-,0)(0,)
Step 9.2
Substitute any number, such as -2, from the interval (-,0) in the first derivative -3x|x| to check if the result is negative or positive.
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Step 9.2.1
Replace the variable x with -2 in the expression.
f(-2)=-3(-2)|-2|
Step 9.2.2
Simplify the result.
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Step 9.2.2.1
Multiply 3 by -2.
f(-2)=--6|-2|
Step 9.2.2.2
The absolute value is the distance between a number and zero. The distance between -2 and 0 is 2.
f(-2)=--62
Step 9.2.2.3
Divide -6 by 2.
f(-2)=3
Step 9.2.2.4
The final answer is 3.
3
3
3
Step 9.3
Substitute any number, such as 2, from the interval (0,) in the first derivative -3x|x| to check if the result is negative or positive.
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Step 9.3.1
Replace the variable x with 2 in the expression.
f(2)=-3(2)|2|
Step 9.3.2
Simplify the result.
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Step 9.3.2.1
Multiply 3 by 2.
f(2)=-6|2|
Step 9.3.2.2
The absolute value is the distance between a number and zero. The distance between 0 and 2 is 2.
f(2)=-62
Step 9.3.2.3
Divide 6 by 2.
f(2)=-13
Step 9.3.2.4
Multiply -1 by 3.
f(2)=-3
Step 9.3.2.5
The final answer is -3.
-3
-3
-3
Step 9.4
Since the first derivative changed signs from positive to negative around x=0, then x=0 is a local maximum.
x=0 is a local maximum
x=0 is a local maximum
Step 10
 [x2  12  π  xdx ]