Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=x^a(1-x)^b , 0<=x<=1
f(x)=xa(1-x)bf(x)=xa(1x)b , 0x1
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 Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=xa and g(x)=(1-x)b.
xaddx[(1-x)b]+(1-x)bddx[xa]
Step 1.1.1.2
Differentiate using the chain rule, which states that ddx[f(g(x))] is f(g(x))g(x) where f(x)=xb and g(x)=1-x.
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Step 1.1.1.2.1
To apply the Chain Rule, set u as 1-x.
xa(ddu[ub]ddx[1-x])+(1-x)bddx[xa]
Step 1.1.1.2.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=b.
xa(bub-1ddx[1-x])+(1-x)bddx[xa]
Step 1.1.1.2.3
Replace all occurrences of u with 1-x.
xa(b(1-x)b-1ddx[1-x])+(1-x)bddx[xa]
xa(b(1-x)b-1ddx[1-x])+(1-x)bddx[xa]
Step 1.1.1.3
Differentiate.
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Step 1.1.1.3.1
By the Sum Rule, the derivative of 1-x with respect to x is ddx[1]+ddx[-x].
xa(b(1-x)b-1(ddx[1]+ddx[-x]))+(1-x)bddx[xa]
Step 1.1.1.3.2
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
xa(b(1-x)b-1(0+ddx[-x]))+(1-x)bddx[xa]
Step 1.1.1.3.3
Add 0 and ddx[-x].
xa(b(1-x)b-1ddx[-x])+(1-x)bddx[xa]
Step 1.1.1.3.4
Since -1 is constant with respect to x, the derivative of -x with respect to x is -ddx[x].
xa(b(1-x)b-1(-ddx[x]))+(1-x)bddx[xa]
Step 1.1.1.3.5
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
xa(b(1-x)b-1(-11))+(1-x)bddx[xa]
Step 1.1.1.3.6
Simplify the expression.
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Step 1.1.1.3.6.1
Multiply -1 by 1.
xa(b(1-x)b-1-1)+(1-x)bddx[xa]
Step 1.1.1.3.6.2
Move -1 to the left of b(1-x)b-1.
xa(-1(b(1-x)b-1))+(1-x)bddx[xa]
Step 1.1.1.3.6.3
Rewrite -1b as -b.
xa(-b(1-x)b-1)+(1-x)bddx[xa]
xa(-b(1-x)b-1)+(1-x)bddx[xa]
Step 1.1.1.3.7
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=a.
xa(-b(1-x)b-1)+(1-x)b(axa-1)
xa(-b(1-x)b-1)+(1-x)b(axa-1)
Step 1.1.1.4
Simplify.
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Step 1.1.1.4.1
Reorder terms.
-xab(1-x)b-1+xa-1a(1-x)b
Step 1.1.1.4.2
Reorder factors in -xab(1-x)b-1+xa-1a(1-x)b.
f(x)=-bxa(1-x)b-1+axa-1(1-x)b
f(x)=-bxa(1-x)b-1+axa-1(1-x)b
f(x)=-bxa(1-x)b-1+axa-1(1-x)b
Step 1.1.2
The first derivative of f(x) with respect to x is -bxa(1-x)b-1+axa-1(1-x)b.
-bxa(1-x)b-1+axa-1(1-x)b
-bxa(1-x)b-1+axa-1(1-x)b
Step 1.2
Set the first derivative equal to 0.
-bxa(1-x)b-1+axa-1(1-x)b=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
There are no values of x in the domain of the original problem where the derivative is 0 or undefined.
No critical points found
No critical points found
Step 2
Evaluate at the included endpoints.
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Step 2.1
Evaluate at x=0.
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Step 2.1.1
Substitute 0 for x.
(0)a(1-(0))b
Step 2.1.2
Simplify.
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Step 2.1.2.1
Subtract 0 from 1.
0a1b
Step 2.1.2.2
One to any power is one.
0a1
Step 2.1.2.3
Multiply 0a by 1.
0a
0a
0a
Step 2.2
Evaluate at x=1.
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Step 2.2.1
Substitute 1 for x.
(1)a(1-(1))b
Step 2.2.2
Simplify.
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Step 2.2.2.1
One to any power is one.
1(1-(1))b
Step 2.2.2.2
Multiply (1-(1))b by 1.
(1-(1))b
Step 2.2.2.3
Multiply -1 by 1.
(1-1)b
Step 2.2.2.4
Subtract 1 from 1.
0b
0b
0b
Step 2.3
List all of the points.
(0,0a),(1,0b)
(0,0a),(1,0b)
Step 3
Compare the f(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 f(x) value and the minimum will occur at the lowest f(x) value.
No absolute maximum
No absolute minimum
Step 4
 [x2  12  π  xdx ]