Calculus Examples

Find the Absolute Max and Min over the Interval 1/x
1x
Step 1
Find the first derivative of the function.
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Step 1.1
Rewrite 1x as x-1.
ddx[x-1]
Step 1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=-1.
-x-2
Step 1.3
Rewrite the expression using the negative exponent rule b-n=1bn.
-1x2
-1x2
Step 2
Find the second derivative of the function.
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Step 2.1
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)=-1 and g(x)=1x2.
f′′(x)=-ddx1x2+1x2ddx(-1)
Step 2.2
Differentiate.
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Step 2.2.1
Rewrite 1x2 as (x2)-1.
f′′(x)=-ddx(x2)-1+1x2ddx(-1)
Step 2.2.2
Multiply the exponents in (x2)-1.
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Step 2.2.2.1
Apply the power rule and multiply exponents, (am)n=amn.
f′′(x)=-ddxx2-1+1x2ddx(-1)
Step 2.2.2.2
Multiply 2 by -1.
f′′(x)=-ddxx-2+1x2ddx(-1)
f′′(x)=-ddxx-2+1x2ddx(-1)
Step 2.2.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=-2.
f′′(x)=-(-2x-3)+1x2ddx(-1)
Step 2.2.4
Multiply -2 by -1.
f′′(x)=2x-3+1x2ddx(-1)
Step 2.2.5
Since -1 is constant with respect to x, the derivative of -1 with respect to x is 0.
f′′(x)=2x-3+1x20
Step 2.2.6
Simplify the expression.
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Step 2.2.6.1
Multiply 1x2 by 0.
f′′(x)=2x-3+0
Step 2.2.6.2
Add 2x-3 and 0.
f′′(x)=2x-3
f′′(x)=2x-3
f′′(x)=2x-3
Step 2.3
Simplify.
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Step 2.3.1
Rewrite the expression using the negative exponent rule b-n=1bn.
f′′(x)=2(1x3)
Step 2.3.2
Combine 2 and 1x3.
f′′(x)=2x3
f′′(x)=2x3
f′′(x)=2x3
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
-1x2=0
Step 4
Since there is no value of x that makes the first derivative equal to 0, there are no local extrema.
No Local Extrema
Step 5
No Local Extrema
Step 6
 [x2  12  π  xdx ]