Calculus Examples

Find the Absolute Max and Min over the Interval g(x)=e^(-x^4) , -2<=x<=1
g(x)=e-x4g(x)=ex4 , -2x12x1
Step 1
Find the critical points.
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Step 1.1
Find the first derivative.
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Step 1.1.1
Find the first derivative.
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Step 1.1.1.1
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f(g(x))g(x) where f(x)=ex and g(x)=-x4.
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Step 1.1.1.1.1
To apply the Chain Rule, set u as -x4.
ddu[eu]ddx[-x4]
Step 1.1.1.1.2
Differentiate using the Exponential Rule which states that ddu[au] is auln(a) where a=e.
euddx[-x4]
Step 1.1.1.1.3
Replace all occurrences of u with -x4.
e-x4ddx[-x4]
e-x4ddx[-x4]
Step 1.1.1.2
Differentiate.
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Step 1.1.1.2.1
Since -1 is constant with respect to x, the derivative of -x4 with respect to x is -ddx[x4].
e-x4(-ddx[x4])
Step 1.1.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=4.
e-x4(-(4x3))
Step 1.1.1.2.3
Multiply 4 by -1.
e-x4(-4x3)
e-x4(-4x3)
Step 1.1.1.3
Simplify.
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Step 1.1.1.3.1
Reorder the factors of e-x4(-4x3).
-4e-x4x3
Step 1.1.1.3.2
Reorder factors in -4e-x4x3.
f(x)=-4x3e-x4
f(x)=-4x3e-x4
f(x)=-4x3e-x4
Step 1.1.2
The first derivative of g(x) with respect to x is -4x3e-x4.
-4x3e-x4
-4x3e-x4
Step 1.2
Set the first derivative equal to 0 then solve the equation -4x3e-x4=0.
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Step 1.2.1
Set the first derivative equal to 0.
-4x3e-x4=0
Step 1.2.2
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
x3=0
e-x4=0
Step 1.2.3
Set x3 equal to 0 and solve for x.
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Step 1.2.3.1
Set x3 equal to 0.
x3=0
Step 1.2.3.2
Solve x3=0 for x.
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Step 1.2.3.2.1
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=30
Step 1.2.3.2.2
Simplify 30.
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Step 1.2.3.2.2.1
Rewrite 0 as 03.
x=303
Step 1.2.3.2.2.2
Pull terms out from under the radical, assuming real numbers.
x=0
x=0
x=0
x=0
Step 1.2.4
Set e-x4 equal to 0 and solve for x.
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Step 1.2.4.1
Set e-x4 equal to 0.
e-x4=0
Step 1.2.4.2
Solve e-x4=0 for x.
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Step 1.2.4.2.1
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(e-x4)=ln(0)
Step 1.2.4.2.2
The equation cannot be solved because ln(0) is undefined.
Undefined
Step 1.2.4.2.3
There is no solution for e-x4=0
No solution
No solution
No solution
Step 1.2.5
The final solution is all the values that make -4x3e-x4=0 true.
x=0
x=0
Step 1.3
Find the values where the derivative is undefined.
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Step 1.3.1
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Step 1.4
Evaluate e-x4 at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=0.
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Step 1.4.1.1
Substitute 0 for x.
e-(0)4
Step 1.4.1.2
Simplify.
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Step 1.4.1.2.1
Raising 0 to any positive power yields 0.
e-0
Step 1.4.1.2.2
Multiply -1 by 0.
e0
Step 1.4.1.2.3
Anything raised to 0 is 1.
1
1
1
Step 1.4.2
List all of the points.
(0,1)
(0,1)
(0,1)
Step 2
Evaluate at the included endpoints.
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Step 2.1
Evaluate at x=-2.
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Step 2.1.1
Substitute -2 for x.
e-(-2)4
Step 2.1.2
Simplify.
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Step 2.1.2.1
Raise -2 to the power of 4.
e-116
Step 2.1.2.2
Multiply -1 by 16.
e-16
Step 2.1.2.3
Rewrite the expression using the negative exponent rule b-n=1bn.
1e16
1e16
1e16
Step 2.2
Evaluate at x=1.
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Step 2.2.1
Substitute 1 for x.
e-(1)4
Step 2.2.2
Simplify.
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Step 2.2.2.1
One to any power is one.
e-11
Step 2.2.2.2
Multiply -1 by 1.
e-1
Step 2.2.2.3
Rewrite the expression using the negative exponent rule b-n=1bn.
1e
1e
1e
Step 2.3
List all of the points.
(-2,1e16),(1,1e)
(-2,1e16),(1,1e)
Step 3
Compare the g(x) values found for each value of x in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest g(x) value and the minimum will occur at the lowest g(x) value.
Absolute Maximum: (0,1)
Absolute Minimum: (-2,1e16)
Step 4
 [x2  12  π  xdx ]