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Calculus Examples
f(x)=9-x2f(x)=9−x2
Step 1
Step 1.1
Differentiate.
Step 1.1.1
By the Sum Rule, the derivative of 9-x29−x2 with respect to xx is ddx[9]+ddx[-x2]ddx[9]+ddx[−x2].
ddx[9]+ddx[-x2]ddx[9]+ddx[−x2]
Step 1.1.2
Since 99 is constant with respect to xx, the derivative of 99 with respect to xx is 00.
0+ddx[-x2]0+ddx[−x2]
0+ddx[-x2]0+ddx[−x2]
Step 1.2
Evaluate ddx[-x2]ddx[−x2].
Step 1.2.1
Since -1−1 is constant with respect to xx, the derivative of -x2−x2 with respect to x is -ddx[x2].
0-ddx[x2]
Step 1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
0-(2x)
Step 1.2.3
Multiply 2 by -1.
0-2x
0-2x
Step 1.3
Subtract 2x from 0.
-2x
-2x
Step 2
Step 2.1
Since -2 is constant with respect to x, the derivative of -2x with respect to x is -2ddx[x].
f′′(x)=-2ddxx
Step 2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
f′′(x)=-2⋅1
Step 2.3
Multiply -2 by 1.
f′′(x)=-2
f′′(x)=-2
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
-2x=0
Step 4
Step 4.1
Find the first derivative.
Step 4.1.1
Differentiate.
Step 4.1.1.1
By the Sum Rule, the derivative of 9-x2 with respect to x is ddx[9]+ddx[-x2].
ddx[9]+ddx[-x2]
Step 4.1.1.2
Since 9 is constant with respect to x, the derivative of 9 with respect to x is 0.
0+ddx[-x2]
0+ddx[-x2]
Step 4.1.2
Evaluate ddx[-x2].
Step 4.1.2.1
Since -1 is constant with respect to x, the derivative of -x2 with respect to x is -ddx[x2].
0-ddx[x2]
Step 4.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
0-(2x)
Step 4.1.2.3
Multiply 2 by -1.
0-2x
0-2x
Step 4.1.3
Subtract 2x from 0.
f′(x)=-2x
f′(x)=-2x
Step 4.2
The first derivative of f(x) with respect to x is -2x.
-2x
-2x
Step 5
Step 5.1
Set the first derivative equal to 0.
-2x=0
Step 5.2
Divide each term in -2x=0 by -2 and simplify.
Step 5.2.1
Divide each term in -2x=0 by -2.
-2x-2=0-2
Step 5.2.2
Simplify the left side.
Step 5.2.2.1
Cancel the common factor of -2.
Step 5.2.2.1.1
Cancel the common factor.
-2x-2=0-2
Step 5.2.2.1.2
Divide x by 1.
x=0-2
x=0-2
x=0-2
Step 5.2.3
Simplify the right side.
Step 5.2.3.1
Divide 0 by -2.
x=0
x=0
x=0
x=0
Step 6
Step 6.1
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 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.
-2
Step 9
x=0 is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
x=0 is a local maximum
Step 10
Step 10.1
Replace the variable x with 0 in the expression.
f(0)=9-(0)2
Step 10.2
Simplify the result.
Step 10.2.1
Simplify each term.
Step 10.2.1.1
Raising 0 to any positive power yields 0.
f(0)=9-0
Step 10.2.1.2
Multiply -1 by 0.
f(0)=9+0
f(0)=9+0
Step 10.2.2
Add 9 and 0.
f(0)=9
Step 10.2.3
The final answer is 9.
y=9
y=9
y=9
Step 11
These are the local extrema for f(x)=9-x2.
(0,9) is a local maxima
Step 12