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Calculus Examples
y=2-2x-x3y=2−2x−x3
Step 1
Write y=2-2x-x3 as a function.
f(x)=2-2x-x3
Step 2
Step 2.1
Find the first derivative.
Step 2.1.1
Differentiate.
Step 2.1.1.1
By the Sum Rule, the derivative of 2-2x-x3 with respect to x is ddx[2]+ddx[-2x]+ddx[-x3].
ddx[2]+ddx[-2x]+ddx[-x3]
Step 2.1.1.2
Since 2 is constant with respect to x, the derivative of 2 with respect to x is 0.
0+ddx[-2x]+ddx[-x3]
0+ddx[-2x]+ddx[-x3]
Step 2.1.2
Evaluate ddx[-2x].
Step 2.1.2.1
Since -2 is constant with respect to x, the derivative of -2x with respect to x is -2ddx[x].
0-2ddx[x]+ddx[-x3]
Step 2.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
0-2⋅1+ddx[-x3]
Step 2.1.2.3
Multiply -2 by 1.
0-2+ddx[-x3]
0-2+ddx[-x3]
Step 2.1.3
Evaluate ddx[-x3].
Step 2.1.3.1
Since -1 is constant with respect to x, the derivative of -x3 with respect to x is -ddx[x3].
0-2-ddx[x3]
Step 2.1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
0-2-(3x2)
Step 2.1.3.3
Multiply 3 by -1.
0-2-3x2
0-2-3x2
Step 2.1.4
Simplify.
Step 2.1.4.1
Subtract 2 from 0.
-2-3x2
Step 2.1.4.2
Reorder terms.
f′(x)=-3x2-2
f′(x)=-3x2-2
f′(x)=-3x2-2
Step 2.2
The first derivative of f(x) with respect to x is -3x2-2.
-3x2-2
-3x2-2
Step 3
Step 3.1
Set the first derivative equal to 0.
-3x2-2=0
Step 3.2
Add 2 to both sides of the equation.
-3x2=2
Step 3.3
Divide each term in -3x2=2 by -3 and simplify.
Step 3.3.1
Divide each term in -3x2=2 by -3.
-3x2-3=2-3
Step 3.3.2
Simplify the left side.
Step 3.3.2.1
Cancel the common factor of -3.
Step 3.3.2.1.1
Cancel the common factor.
-3x2-3=2-3
Step 3.3.2.1.2
Divide x2 by 1.
x2=2-3
x2=2-3
x2=2-3
Step 3.3.3
Simplify the right side.
Step 3.3.3.1
Move the negative in front of the fraction.
x2=-23
x2=-23
x2=-23
Step 3.4
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±√-23
Step 3.5
Simplify ±√-23.
Step 3.5.1
Rewrite -1 as i2.
x=±√i223
Step 3.5.2
Pull terms out from under the radical.
x=±i√23
Step 3.5.3
Rewrite √23 as √2√3.
x=±i√2√3
Step 3.5.4
Multiply √2√3 by √3√3.
x=±i(√2√3⋅√3√3)
Step 3.5.5
Combine and simplify the denominator.
Step 3.5.5.1
Multiply √2√3 by √3√3.
x=±i√2√3√3√3
Step 3.5.5.2
Raise √3 to the power of 1.
x=±i√2√3√31√3
Step 3.5.5.3
Raise √3 to the power of 1.
x=±i√2√3√31√31
Step 3.5.5.4
Use the power rule aman=am+n to combine exponents.
x=±i√2√3√31+1
Step 3.5.5.5
Add 1 and 1.
x=±i√2√3√32
Step 3.5.5.6
Rewrite √32 as 3.
Step 3.5.5.6.1
Use n√ax=axn to rewrite √3 as 312.
x=±i√2√3(312)2
Step 3.5.5.6.2
Apply the power rule and multiply exponents, (am)n=amn.
x=±i√2√3312⋅2
Step 3.5.5.6.3
Combine 12 and 2.
x=±i√2√3322
Step 3.5.5.6.4
Cancel the common factor of 2.
Step 3.5.5.6.4.1
Cancel the common factor.
x=±i√2√3322
Step 3.5.5.6.4.2
Rewrite the expression.
x=±i√2√331
x=±i√2√331
Step 3.5.5.6.5
Evaluate the exponent.
x=±i√2√33
x=±i√2√33
x=±i√2√33
Step 3.5.6
Simplify the numerator.
Step 3.5.6.1
Combine using the product rule for radicals.
x=±i√2⋅33
Step 3.5.6.2
Multiply 2 by 3.
x=±i√63
x=±i√63
Step 3.5.7
Combine i and √63.
x=±i√63
x=±i√63
Step 3.6
The complete solution is the result of both the positive and negative portions of the solution.
Step 3.6.1
First, use the positive value of the ± to find the first solution.
x=i√63
Step 3.6.2
Next, use the negative value of the ± to find the second solution.
x=-i√63
Step 3.6.3
The complete solution is the result of both the positive and negative portions of the solution.
x=i√63,-i√63
x=i√63,-i√63
x=i√63,-i√63
Step 4
There are no values of x in the domain of the original problem where the derivative is 0 or undefined.
No critical points found
Step 5
No points make the derivative f′(x)=-3x2-2 equal to 0 or undefined. The interval to check if f(x)=2-2x-x3 is increasing or decreasing is (-∞,∞).
(-∞,∞)
Step 6
Step 6.1
Replace the variable x with 1 in the expression.
f′(1)=-3(1)2-2
Step 6.2
Simplify the result.
Step 6.2.1
Simplify each term.
Step 6.2.1.1
One to any power is one.
f′(1)=-3⋅1-2
Step 6.2.1.2
Multiply -3 by 1.
f′(1)=-3-2
f′(1)=-3-2
Step 6.2.2
Subtract 2 from -3.
f′(1)=-5
Step 6.2.3
The final answer is -5.
-5
-5
-5
Step 7
The result of substituting 1 into f′(x)=-3x2-2 is -5, which is negative, so the graph is decreasing on the interval (-∞,∞).
Decreasing on (-∞,∞)
Step 8
Decreasing over the interval (-∞,∞) means that the function is always decreasing.
Always Decreasing
Step 9