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Calculus Examples
f(x)=xx2+25f(x)=xx2+25 on -7−7 , 77
Step 1
Step 1.1
Find the first derivative.
Step 1.1.1
Find the first derivative.
Step 1.1.1.1
Differentiate using the Quotient Rule which states that ddx[f(x)g(x)]ddx[f(x)g(x)] is g(x)ddx[f(x)]-f(x)ddx[g(x)]g(x)2g(x)ddx[f(x)]−f(x)ddx[g(x)]g(x)2 where f(x)=xf(x)=x and g(x)=x2+25g(x)=x2+25.
(x2+25)ddx[x]-xddx[x2+25](x2+25)2(x2+25)ddx[x]−xddx[x2+25](x2+25)2
Step 1.1.1.2
Differentiate.
Step 1.1.1.2.1
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
(x2+25)⋅1-xddx[x2+25](x2+25)2(x2+25)⋅1−xddx[x2+25](x2+25)2
Step 1.1.1.2.2
Multiply x2+25x2+25 by 11.
x2+25-xddx[x2+25](x2+25)2x2+25−xddx[x2+25](x2+25)2
Step 1.1.1.2.3
By the Sum Rule, the derivative of x2+25x2+25 with respect to xx is ddx[x2]+ddx[25]ddx[x2]+ddx[25].
x2+25-x(ddx[x2]+ddx[25])(x2+25)2x2+25−x(ddx[x2]+ddx[25])(x2+25)2
Step 1.1.1.2.4
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=2n=2.
x2+25-x(2x+ddx[25])(x2+25)2x2+25−x(2x+ddx[25])(x2+25)2
Step 1.1.1.2.5
Since 2525 is constant with respect to xx, the derivative of 2525 with respect to xx is 00.
x2+25-x(2x+0)(x2+25)2x2+25−x(2x+0)(x2+25)2
Step 1.1.1.2.6
Simplify the expression.
Step 1.1.1.2.6.1
Add 2x2x and 00.
x2+25-x(2x)(x2+25)2x2+25−x(2x)(x2+25)2
Step 1.1.1.2.6.2
Multiply 22 by -1−1.
x2+25-2x⋅x(x2+25)2x2+25−2x⋅x(x2+25)2
x2+25-2x⋅x(x2+25)2x2+25−2x⋅x(x2+25)2
x2+25-2x⋅x(x2+25)2x2+25−2x⋅x(x2+25)2
Step 1.1.1.3
Raise xx to the power of 11.
x2+25-2(x1x)(x2+25)2x2+25−2(x1x)(x2+25)2
Step 1.1.1.4
Raise xx to the power of 11.
x2+25-2(x1x1)(x2+25)2x2+25−2(x1x1)(x2+25)2
Step 1.1.1.5
Use the power rule aman=am+naman=am+n to combine exponents.
x2+25-2x1+1(x2+25)2x2+25−2x1+1(x2+25)2
Step 1.1.1.6
Add 11 and 11.
x2+25-2x2(x2+25)2x2+25−2x2(x2+25)2
Step 1.1.1.7
Subtract 2x22x2 from x2x2.
f′(x)=-x2+25(x2+25)2
f′(x)=-x2+25(x2+25)2
Step 1.1.2
The first derivative of f(x) with respect to x is -x2+25(x2+25)2.
-x2+25(x2+25)2
-x2+25(x2+25)2
Step 1.2
Set the first derivative equal to 0 then solve the equation -x2+25(x2+25)2=0.
Step 1.2.1
Set the first derivative equal to 0.
-x2+25(x2+25)2=0
Step 1.2.2
Set the numerator equal to zero.
-x2+25=0
Step 1.2.3
Solve the equation for x.
Step 1.2.3.1
Subtract 25 from both sides of the equation.
-x2=-25
Step 1.2.3.2
Divide each term in -x2=-25 by -1 and simplify.
Step 1.2.3.2.1
Divide each term in -x2=-25 by -1.
-x2-1=-25-1
Step 1.2.3.2.2
Simplify the left side.
Step 1.2.3.2.2.1
Dividing two negative values results in a positive value.
x21=-25-1
Step 1.2.3.2.2.2
Divide x2 by 1.
x2=-25-1
x2=-25-1
Step 1.2.3.2.3
Simplify the right side.
Step 1.2.3.2.3.1
Divide -25 by -1.
x2=25
x2=25
x2=25
Step 1.2.3.3
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
x=±√25
Step 1.2.3.4
Simplify ±√25.
Step 1.2.3.4.1
Rewrite 25 as 52.
x=±√52
Step 1.2.3.4.2
Pull terms out from under the radical, assuming positive real numbers.
x=±5
x=±5
Step 1.2.3.5
The complete solution is the result of both the positive and negative portions of the solution.
Step 1.2.3.5.1
First, use the positive value of the ± to find the first solution.
x=5
Step 1.2.3.5.2
Next, use the negative value of the ± to find the second solution.
x=-5
Step 1.2.3.5.3
The complete solution is the result of both the positive and negative portions of the solution.
x=5,-5
x=5,-5
x=5,-5
x=5,-5
Step 1.3
Find the values where the derivative is undefined.
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 xx2+25 at each x value where the derivative is 0 or undefined.
Step 1.4.1
Evaluate at x=5.
Step 1.4.1.1
Substitute 5 for x.
5(5)2+25
Step 1.4.1.2
Simplify.
Step 1.4.1.2.1
Simplify the denominator.
Step 1.4.1.2.1.1
Raise 5 to the power of 2.
525+25
Step 1.4.1.2.1.2
Add 25 and 25.
550
550
Step 1.4.1.2.2
Cancel the common factor of 5 and 50.
Step 1.4.1.2.2.1
Factor 5 out of 5.
5(1)50
Step 1.4.1.2.2.2
Cancel the common factors.
Step 1.4.1.2.2.2.1
Factor 5 out of 50.
5⋅15⋅10
Step 1.4.1.2.2.2.2
Cancel the common factor.
5⋅15⋅10
Step 1.4.1.2.2.2.3
Rewrite the expression.
110
110
110
110
110
Step 1.4.2
Evaluate at x=-5.
Step 1.4.2.1
Substitute -5 for x.
-5(-5)2+25
Step 1.4.2.2
Simplify.
Step 1.4.2.2.1
Simplify the denominator.
Step 1.4.2.2.1.1
Raise -5 to the power of 2.
-525+25
Step 1.4.2.2.1.2
Add 25 and 25.
-550
-550
Step 1.4.2.2.2
Reduce the expression by cancelling the common factors.
Step 1.4.2.2.2.1
Cancel the common factor of -5 and 50.
Step 1.4.2.2.2.1.1
Factor 5 out of -5.
5(-1)50
Step 1.4.2.2.2.1.2
Cancel the common factors.
Step 1.4.2.2.2.1.2.1
Factor 5 out of 50.
5⋅-15⋅10
Step 1.4.2.2.2.1.2.2
Cancel the common factor.
5⋅-15⋅10
Step 1.4.2.2.2.1.2.3
Rewrite the expression.
-110
-110
-110
Step 1.4.2.2.2.2
Move the negative in front of the fraction.
-110
-110
-110
-110
Step 1.4.3
List all of the points.
(5,110),(-5,-110)
(5,110),(-5,-110)
(5,110),(-5,-110)
Step 2
Step 2.1
Evaluate at x=-7.
Step 2.1.1
Substitute -7 for x.
-7(-7)2+25
Step 2.1.2
Simplify.
Step 2.1.2.1
Simplify the denominator.
Step 2.1.2.1.1
Raise -7 to the power of 2.
-749+25
Step 2.1.2.1.2
Add 49 and 25.
-774
-774
Step 2.1.2.2
Move the negative in front of the fraction.
-774
-774
-774
Step 2.2
Evaluate at x=7.
Step 2.2.1
Substitute 7 for x.
7(7)2+25
Step 2.2.2
Simplify the denominator.
Step 2.2.2.1
Raise 7 to the power of 2.
749+25
Step 2.2.2.2
Add 49 and 25.
774
774
774
Step 2.3
List all of the points.
(-7,-774),(7,774)
(-7,-774),(7,774)
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.
Absolute Maximum: (5,110)
Absolute Minimum: (-5,-110)
Step 4