Calculus Examples

Find the Absolute Max and Min over the Interval g(x)=-x^2+24x-109 , si*8<=x<=14 ?
g(x)=-x2+24x-109 , si8x14 ?
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
By the Sum Rule, the derivative of -x2+24x-109 with respect to x is ddx[-x2]+ddx[24x]+ddx[-109].
ddx[-x2]+ddx[24x]+ddx[-109]
Step 1.1.1.2
Evaluate ddx[-x2].
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Step 1.1.1.2.1
Since -1 is constant with respect to x, the derivative of -x2 with respect to x is -ddx[x2].
-ddx[x2]+ddx[24x]+ddx[-109]
Step 1.1.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
-(2x)+ddx[24x]+ddx[-109]
Step 1.1.1.2.3
Multiply 2 by -1.
-2x+ddx[24x]+ddx[-109]
-2x+ddx[24x]+ddx[-109]
Step 1.1.1.3
Evaluate ddx[24x].
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Step 1.1.1.3.1
Since 24 is constant with respect to x, the derivative of 24x with respect to x is 24ddx[x].
-2x+24ddx[x]+ddx[-109]
Step 1.1.1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
-2x+241+ddx[-109]
Step 1.1.1.3.3
Multiply 24 by 1.
-2x+24+ddx[-109]
-2x+24+ddx[-109]
Step 1.1.1.4
Differentiate using the Constant Rule.
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Step 1.1.1.4.1
Since -109 is constant with respect to x, the derivative of -109 with respect to x is 0.
-2x+24+0
Step 1.1.1.4.2
Add -2x+24 and 0.
f(x)=-2x+24
f(x)=-2x+24
f(x)=-2x+24
Step 1.1.2
The first derivative of g(x) with respect to x is -2x+24.
-2x+24
-2x+24
Step 1.2
Set the first derivative equal to 0 then solve the equation -2x+24=0.
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Step 1.2.1
Set the first derivative equal to 0.
-2x+24=0
Step 1.2.2
Subtract 24 from both sides of the equation.
-2x=-24
Step 1.2.3
Divide each term in -2x=-24 by -2 and simplify.
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Step 1.2.3.1
Divide each term in -2x=-24 by -2.
-2x-2=-24-2
Step 1.2.3.2
Simplify the left side.
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Step 1.2.3.2.1
Cancel the common factor of -2.
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Step 1.2.3.2.1.1
Cancel the common factor.
-2x-2=-24-2
Step 1.2.3.2.1.2
Divide x by 1.
x=-24-2
x=-24-2
x=-24-2
Step 1.2.3.3
Simplify the right side.
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Step 1.2.3.3.1
Divide -24 by -2.
x=12
x=12
x=12
x=12
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 -x2+24x-109 at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=12.
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Step 1.4.1.1
Substitute 12 for x.
-(12)2+24(12)-109
Step 1.4.1.2
Simplify.
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Step 1.4.1.2.1
Simplify each term.
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Step 1.4.1.2.1.1
Raise 12 to the power of 2.
-1144+24(12)-109
Step 1.4.1.2.1.2
Multiply -1 by 144.
-144+24(12)-109
Step 1.4.1.2.1.3
Multiply 24 by 12.
-144+288-109
-144+288-109
Step 1.4.1.2.2
Simplify by adding and subtracting.
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Step 1.4.1.2.2.1
Add -144 and 288.
144-109
Step 1.4.1.2.2.2
Subtract 109 from 144.
35
35
35
35
Step 1.4.2
List all of the points.
(12,35)
(12,35)
(12,35)
Step 2
Exclude the points that are not on the interval.
Step 3
Evaluate at the included endpoints.
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Step 3.1
Evaluate at x=8si.
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Step 3.1.1
Substitute 8si for x.
-(8si)2+24(8si)-109
Step 3.1.2
Simplify each term.
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Step 3.1.2.1
Use the power rule (ab)n=anbn to distribute the exponent.
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Step 3.1.2.1.1
Apply the product rule to 8si.
-((8s)2i2)+24(8si)-109
Step 3.1.2.1.2
Apply the product rule to 8s.
-(82s2i2)+24(8si)-109
-(82s2i2)+24(8si)-109
Step 3.1.2.2
Raise 8 to the power of 2.
-(64s2i2)+24(8si)-109
Step 3.1.2.3
Rewrite i2 as -1.
-(64s2-1)+24(8si)-109
Step 3.1.2.4
Multiply -1 by 64.
-(-64s2)+24(8si)-109
Step 3.1.2.5
Multiply -64 by -1.
64s2+24(8si)-109
Step 3.1.2.6
Multiply 8 by 24.
64s2+192si-109
64s2+192si-109
64s2+192si-109
Step 3.2
Evaluate at x=14.
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Step 3.2.1
Substitute 14 for x.
-(14)2+24(14)-109
Step 3.2.2
Simplify.
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Step 3.2.2.1
Simplify each term.
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Step 3.2.2.1.1
Raise 14 to the power of 2.
-1196+24(14)-109
Step 3.2.2.1.2
Multiply -1 by 196.
-196+24(14)-109
Step 3.2.2.1.3
Multiply 24 by 14.
-196+336-109
-196+336-109
Step 3.2.2.2
Simplify by adding and subtracting.
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Step 3.2.2.2.1
Add -196 and 336.
140-109
Step 3.2.2.2.2
Subtract 109 from 140.
31
31
31
31
Step 3.3
List all of the points.
(14,31)
(14,31)
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
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: (14,31)
No absolute minimum
Step 6
 [x2  12  π  xdx ]