Calculus Examples

Find the Critical Points f(x)=(x+2)/(x^2-3x-10)
f(x)=x+2x2-3x-10f(x)=x+2x23x10
Step 1
Find the first derivative.
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Step 1.1
Find the first derivative.
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Step 1.1.1
Differentiate using the Quotient Rule which states that ddx[f(x)g(x)] is g(x)ddx[f(x)]-f(x)ddx[g(x)]g(x)2 where f(x)=x+2 and g(x)=x2-3x-10.
(x2-3x-10)ddx[x+2]-(x+2)ddx[x2-3x-10](x2-3x-10)2
Step 1.1.2
Differentiate.
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Step 1.1.2.1
By the Sum Rule, the derivative of x+2 with respect to x is ddx[x]+ddx[2].
(x2-3x-10)(ddx[x]+ddx[2])-(x+2)ddx[x2-3x-10](x2-3x-10)2
Step 1.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
(x2-3x-10)(1+ddx[2])-(x+2)ddx[x2-3x-10](x2-3x-10)2
Step 1.1.2.3
Since 2 is constant with respect to x, the derivative of 2 with respect to x is 0.
(x2-3x-10)(1+0)-(x+2)ddx[x2-3x-10](x2-3x-10)2
Step 1.1.2.4
Simplify the expression.
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Step 1.1.2.4.1
Add 1 and 0.
(x2-3x-10)1-(x+2)ddx[x2-3x-10](x2-3x-10)2
Step 1.1.2.4.2
Multiply x2-3x-10 by 1.
x2-3x-10-(x+2)ddx[x2-3x-10](x2-3x-10)2
x2-3x-10-(x+2)ddx[x2-3x-10](x2-3x-10)2
Step 1.1.2.5
By the Sum Rule, the derivative of x2-3x-10 with respect to x is ddx[x2]+ddx[-3x]+ddx[-10].
x2-3x-10-(x+2)(ddx[x2]+ddx[-3x]+ddx[-10])(x2-3x-10)2
Step 1.1.2.6
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
x2-3x-10-(x+2)(2x+ddx[-3x]+ddx[-10])(x2-3x-10)2
Step 1.1.2.7
Since -3 is constant with respect to x, the derivative of -3x with respect to x is -3ddx[x].
x2-3x-10-(x+2)(2x-3ddx[x]+ddx[-10])(x2-3x-10)2
Step 1.1.2.8
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
x2-3x-10-(x+2)(2x-31+ddx[-10])(x2-3x-10)2
Step 1.1.2.9
Multiply -3 by 1.
x2-3x-10-(x+2)(2x-3+ddx[-10])(x2-3x-10)2
Step 1.1.2.10
Since -10 is constant with respect to x, the derivative of -10 with respect to x is 0.
x2-3x-10-(x+2)(2x-3+0)(x2-3x-10)2
Step 1.1.2.11
Add 2x-3 and 0.
x2-3x-10-(x+2)(2x-3)(x2-3x-10)2
x2-3x-10-(x+2)(2x-3)(x2-3x-10)2
Step 1.1.3
Simplify.
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Step 1.1.3.1
Apply the distributive property.
x2-3x-10+(-x-12)(2x-3)(x2-3x-10)2
Step 1.1.3.2
Simplify the numerator.
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Step 1.1.3.2.1
Simplify each term.
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Step 1.1.3.2.1.1
Multiply -1 by 2.
x2-3x-10+(-x-2)(2x-3)(x2-3x-10)2
Step 1.1.3.2.1.2
Expand (-x-2)(2x-3) using the FOIL Method.
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Step 1.1.3.2.1.2.1
Apply the distributive property.
x2-3x-10-x(2x-3)-2(2x-3)(x2-3x-10)2
Step 1.1.3.2.1.2.2
Apply the distributive property.
x2-3x-10-x(2x)-x-3-2(2x-3)(x2-3x-10)2
Step 1.1.3.2.1.2.3
Apply the distributive property.
x2-3x-10-x(2x)-x-3-2(2x)-2-3(x2-3x-10)2
x2-3x-10-x(2x)-x-3-2(2x)-2-3(x2-3x-10)2
Step 1.1.3.2.1.3
Simplify and combine like terms.
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Step 1.1.3.2.1.3.1
Simplify each term.
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Step 1.1.3.2.1.3.1.1
Rewrite using the commutative property of multiplication.
x2-3x-10-12xx-x-3-2(2x)-2-3(x2-3x-10)2
Step 1.1.3.2.1.3.1.2
Multiply x by x by adding the exponents.
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Step 1.1.3.2.1.3.1.2.1
Move x.
x2-3x-10-12(xx)-x-3-2(2x)-2-3(x2-3x-10)2
Step 1.1.3.2.1.3.1.2.2
Multiply x by x.
x2-3x-10-12x2-x-3-2(2x)-2-3(x2-3x-10)2
x2-3x-10-12x2-x-3-2(2x)-2-3(x2-3x-10)2
Step 1.1.3.2.1.3.1.3
Multiply -1 by 2.
x2-3x-10-2x2-x-3-2(2x)-2-3(x2-3x-10)2
Step 1.1.3.2.1.3.1.4
Multiply -3 by -1.
x2-3x-10-2x2+3x-2(2x)-2-3(x2-3x-10)2
Step 1.1.3.2.1.3.1.5
Multiply 2 by -2.
x2-3x-10-2x2+3x-4x-2-3(x2-3x-10)2
Step 1.1.3.2.1.3.1.6
Multiply -2 by -3.
x2-3x-10-2x2+3x-4x+6(x2-3x-10)2
x2-3x-10-2x2+3x-4x+6(x2-3x-10)2
Step 1.1.3.2.1.3.2
Subtract 4x from 3x.
x2-3x-10-2x2-x+6(x2-3x-10)2
x2-3x-10-2x2-x+6(x2-3x-10)2
x2-3x-10-2x2-x+6(x2-3x-10)2
Step 1.1.3.2.2
Subtract 2x2 from x2.
-x2-3x-10-x+6(x2-3x-10)2
Step 1.1.3.2.3
Subtract x from -3x.
-x2-4x-10+6(x2-3x-10)2
Step 1.1.3.2.4
Add -10 and 6.
-x2-4x-4(x2-3x-10)2
-x2-4x-4(x2-3x-10)2
Step 1.1.3.3
Factor by grouping.
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Step 1.1.3.3.1
For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is ac=-1-4=4 and whose sum is b=-4.
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Step 1.1.3.3.1.1
Factor -4 out of -4x.
-x2-4(x)-4(x2-3x-10)2
Step 1.1.3.3.1.2
Rewrite -4 as -2 plus -2
-x2+(-2-2)x-4(x2-3x-10)2
Step 1.1.3.3.1.3
Apply the distributive property.
-x2-2x-2x-4(x2-3x-10)2
-x2-2x-2x-4(x2-3x-10)2
Step 1.1.3.3.2
Factor out the greatest common factor from each group.
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Step 1.1.3.3.2.1
Group the first two terms and the last two terms.
(-x2-2x)-2x-4(x2-3x-10)2
Step 1.1.3.3.2.2
Factor out the greatest common factor (GCF) from each group.
x(-x-2)+2(-x-2)(x2-3x-10)2
x(-x-2)+2(-x-2)(x2-3x-10)2
Step 1.1.3.3.3
Factor the polynomial by factoring out the greatest common factor, -x-2.
(-x-2)(x+2)(x2-3x-10)2
(-x-2)(x+2)(x2-3x-10)2
Step 1.1.3.4
Simplify the denominator.
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Step 1.1.3.4.1
Factor x2-3x-10 using the AC method.
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Step 1.1.3.4.1.1
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -10 and whose sum is -3.
-5,2
Step 1.1.3.4.1.2
Write the factored form using these integers.
(-x-2)(x+2)((x-5)(x+2))2
(-x-2)(x+2)((x-5)(x+2))2
Step 1.1.3.4.2
Apply the product rule to (x-5)(x+2).
(-x-2)(x+2)(x-5)2(x+2)2
(-x-2)(x+2)(x-5)2(x+2)2
Step 1.1.3.5
Simplify the numerator.
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Step 1.1.3.5.1
Factor -1 out of -x.
(-(x)-2)(x+2)(x-5)2(x+2)2
Step 1.1.3.5.2
Rewrite -2 as -1(2).
(-(x)-1(2))(x+2)(x-5)2(x+2)2
Step 1.1.3.5.3
Factor -1 out of -(x)-1(2).
-(x+2)(x+2)(x-5)2(x+2)2
Step 1.1.3.5.4
Rewrite -(x+2) as -1(x+2).
-1(x+2)(x+2)(x-5)2(x+2)2
Step 1.1.3.5.5
Raise x+2 to the power of 1.
-1((x+2)1(x+2))(x-5)2(x+2)2
Step 1.1.3.5.6
Raise x+2 to the power of 1.
-1((x+2)1(x+2)1)(x-5)2(x+2)2
Step 1.1.3.5.7
Use the power rule aman=am+n to combine exponents.
-1(x+2)1+1(x-5)2(x+2)2
Step 1.1.3.5.8
Add 1 and 1.
-1(x+2)2(x-5)2(x+2)2
-1(x+2)2(x-5)2(x+2)2
Step 1.1.3.6
Cancel the common factor of (x+2)2.
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Step 1.1.3.6.1
Cancel the common factor.
-1(x+2)2(x-5)2(x+2)2
Step 1.1.3.6.2
Rewrite the expression.
-1(x-5)2
-1(x-5)2
Step 1.1.3.7
Move the negative in front of the fraction.
f(x)=-1(x-5)2
f(x)=-1(x-5)2
f(x)=-1(x-5)2
Step 1.2
The first derivative of f(x) with respect to x is -1(x-5)2.
-1(x-5)2
-1(x-5)2
Step 2
Set the first derivative equal to 0 then solve the equation -1(x-5)2=0.
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Step 2.1
Set the first derivative equal to 0.
-1(x-5)2=0
Step 2.2
Set the numerator equal to zero.
1=0
Step 2.3
Since 10, there are no solutions.
No solution
No solution
Step 3
Find the values where the derivative is undefined.
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Step 3.1
Set the denominator in 1(x-5)2 equal to 0 to find where the expression is undefined.
(x-5)2=0
Step 3.2
Solve for x.
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Step 3.2.1
Set the x-5 equal to 0.
x-5=0
Step 3.2.2
Add 5 to both sides of the equation.
x=5
x=5
x=5
Step 4
Evaluate x+2x2-3x-10 at each x value where the derivative is 0 or undefined.
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Step 4.1
Evaluate at x=5.
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Step 4.1.1
Substitute 5 for x.
(5)+2(5)2-35-10
Step 4.1.2
Simplify.
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Step 4.1.2.1
Simplify each term.
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Step 4.1.2.1.1
Raise 5 to the power of 2.
(5)+225-35-10
Step 4.1.2.1.2
Multiply -3 by 5.
(5)+225-15-10
(5)+225-15-10
Step 4.1.2.2
Simplify by subtracting numbers.
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Step 4.1.2.2.1
Subtract 15 from 25.
(5)+210-10
Step 4.1.2.2.2
Subtract 10 from 10.
(5)+20
Step 4.1.2.2.3
The expression contains a division by 0. The expression is undefined.
Undefined
(5)+20
Step 4.1.2.3
The expression contains a division by 0. The expression is undefined.
Undefined
Undefined
Undefined
Undefined
Step 5
There are no values of x in the domain of the original problem where the derivative is 0 or undefined.
No critical points found
 [x2  12  π  xdx ]