Calculus Examples
7x2+3x7x2+3x , (1,10)(1,10)
Step 1
Write 7x2+3x7x2+3x as a function.
f(x)=7x2+3xf(x)=7x2+3x
Step 2
Step 2.1
Evaluate f(x)=7x2+3xf(x)=7x2+3x at x=1x=1.
Step 2.1.1
Replace the variable xx with 11 in the expression.
f(1)=7(1)2+3(1)f(1)=7(1)2+3(1)
Step 2.1.2
Simplify the result.
Step 2.1.2.1
Simplify each term.
Step 2.1.2.1.1
One to any power is one.
f(1)=7⋅1+3(1)f(1)=7⋅1+3(1)
Step 2.1.2.1.2
Multiply 77 by 11.
f(1)=7+3(1)f(1)=7+3(1)
Step 2.1.2.1.3
Multiply 33 by 11.
f(1)=7+3f(1)=7+3
f(1)=7+3f(1)=7+3
Step 2.1.2.2
Add 77 and 33.
f(1)=10f(1)=10
Step 2.1.2.3
The final answer is 1010.
1010
1010
1010
Step 2.2
Since 10=1010=10, the point is on the graph.
The point is on the graph
The point is on the graph
Step 3
The slope of the tangent line is the derivative of the expression.
mm == The derivative of f(x)=7x2+3xf(x)=7x2+3x
Step 4
Consider the limit definition of the derivative.
f′(x)=limh→0f(x+h)-f(x)hf'(x)=limh→0f(x+h)−f(x)h
Step 5
Step 5.1
Evaluate the function at x=x+hx=x+h.
Step 5.1.1
Replace the variable xx with x+hx+h in the expression.
f(x+h)=7(x+h)2+3(x+h)f(x+h)=7(x+h)2+3(x+h)
Step 5.1.2
Simplify the result.
Step 5.1.2.1
Simplify each term.
Step 5.1.2.1.1
Rewrite (x+h)2(x+h)2 as (x+h)(x+h)(x+h)(x+h).
f(x+h)=7((x+h)(x+h))+3(x+h)f(x+h)=7((x+h)(x+h))+3(x+h)
Step 5.1.2.1.2
Expand (x+h)(x+h)(x+h)(x+h) using the FOIL Method.
Step 5.1.2.1.2.1
Apply the distributive property.
f(x+h)=7(x(x+h)+h(x+h))+3(x+h)f(x+h)=7(x(x+h)+h(x+h))+3(x+h)
Step 5.1.2.1.2.2
Apply the distributive property.
f(x+h)=7(x⋅x+xh+h(x+h))+3(x+h)f(x+h)=7(x⋅x+xh+h(x+h))+3(x+h)
Step 5.1.2.1.2.3
Apply the distributive property.
f(x+h)=7(x⋅x+xh+hx+h⋅h)+3(x+h)f(x+h)=7(x⋅x+xh+hx+h⋅h)+3(x+h)
f(x+h)=7(x⋅x+xh+hx+h⋅h)+3(x+h)f(x+h)=7(x⋅x+xh+hx+h⋅h)+3(x+h)
Step 5.1.2.1.3
Simplify and combine like terms.
Step 5.1.2.1.3.1
Simplify each term.
Step 5.1.2.1.3.1.1
Multiply xx by xx.
f(x+h)=7(x2+xh+hx+h⋅h)+3(x+h)f(x+h)=7(x2+xh+hx+h⋅h)+3(x+h)
Step 5.1.2.1.3.1.2
Multiply hh by hh.
f(x+h)=7(x2+xh+hx+h2)+3(x+h)f(x+h)=7(x2+xh+hx+h2)+3(x+h)
f(x+h)=7(x2+xh+hx+h2)+3(x+h)f(x+h)=7(x2+xh+hx+h2)+3(x+h)
Step 5.1.2.1.3.2
Add xhxh and hxhx.
Step 5.1.2.1.3.2.1
Reorder xx and hh.
f(x+h)=7(x2+hx+hx+h2)+3(x+h)f(x+h)=7(x2+hx+hx+h2)+3(x+h)
Step 5.1.2.1.3.2.2
Add hxhx and hxhx.
f(x+h)=7(x2+2hx+h2)+3(x+h)f(x+h)=7(x2+2hx+h2)+3(x+h)
f(x+h)=7(x2+2hx+h2)+3(x+h)f(x+h)=7(x2+2hx+h2)+3(x+h)
f(x+h)=7(x2+2hx+h2)+3(x+h)f(x+h)=7(x2+2hx+h2)+3(x+h)
Step 5.1.2.1.4
Apply the distributive property.
f(x+h)=7x2+7(2hx)+7h2+3(x+h)f(x+h)=7x2+7(2hx)+7h2+3(x+h)
Step 5.1.2.1.5
Multiply 22 by 77.
f(x+h)=7x2+14(hx)+7h2+3(x+h)f(x+h)=7x2+14(hx)+7h2+3(x+h)
Step 5.1.2.1.6
Apply the distributive property.
f(x+h)=7x2+14hx+7h2+3x+3hf(x+h)=7x2+14hx+7h2+3x+3h
f(x+h)=7x2+14hx+7h2+3x+3hf(x+h)=7x2+14hx+7h2+3x+3h
Step 5.1.2.2
The final answer is 7x2+14hx+7h2+3x+3h7x2+14hx+7h2+3x+3h.
7x2+14hx+7h2+3x+3h7x2+14hx+7h2+3x+3h
7x2+14hx+7h2+3x+3h7x2+14hx+7h2+3x+3h
7x2+14hx+7h2+3x+3h7x2+14hx+7h2+3x+3h
Step 5.2
Reorder.
Step 5.2.1
Move 3x3x.
7x2+14hx+7h2+3h+3x7x2+14hx+7h2+3h+3x
Step 5.2.2
Move 7x27x2.
14hx+7h2+7x2+3h+3x14hx+7h2+7x2+3h+3x
Step 5.2.3
Reorder 14hx14hx and 7h27h2.
7h2+14hx+7x2+3h+3x7h2+14hx+7x2+3h+3x
7h2+14hx+7x2+3h+3x7h2+14hx+7x2+3h+3x
Step 5.3
Find the components of the definition.
f(x+h)=7h2+14hx+7x2+3h+3xf(x+h)=7h2+14hx+7x2+3h+3x
f(x)=7x2+3xf(x)=7x2+3x
f(x+h)=7h2+14hx+7x2+3h+3xf(x+h)=7h2+14hx+7x2+3h+3x
f(x)=7x2+3xf(x)=7x2+3x
Step 6
Plug in the components.
f′(x)=limh→07h2+14hx+7x2+3h+3x-(7x2+3x)hf'(x)=limh→07h2+14hx+7x2+3h+3x−(7x2+3x)h
Step 7
Step 7.1
Simplify the numerator.
Step 7.1.1
Apply the distributive property.
f′(x)=limh→07h2+14hx+7x2+3h+3x-(7x2)-(3x)hf'(x)=limh→07h2+14hx+7x2+3h+3x−(7x2)−(3x)h
Step 7.1.2
Multiply 77 by -1−1.
f′(x)=limh→07h2+14hx+7x2+3h+3x-7x2-(3x)hf'(x)=limh→07h2+14hx+7x2+3h+3x−7x2−(3x)h
Step 7.1.3
Multiply 33 by -1−1.
f′(x)=limh→07h2+14hx+7x2+3h+3x-7x2-3xhf'(x)=limh→07h2+14hx+7x2+3h+3x−7x2−3xh
Step 7.1.4
Subtract 7x27x2 from 7x27x2.
f′(x)=limh→07h2+14hx+3h+3x+0-3xhf'(x)=limh→07h2+14hx+3h+3x+0−3xh
Step 7.1.5
Add 7h27h2 and 00.
f′(x)=limh→07h2+14hx+3h+3x-3xhf'(x)=limh→07h2+14hx+3h+3x−3xh
Step 7.1.6
Subtract 3x3x from 3x3x.
f′(x)=limh→07h2+14hx+3h+0hf'(x)=limh→07h2+14hx+3h+0h
Step 7.1.7
Add 7h2+14hx+3h7h2+14hx+3h and 00.
f′(x)=limh→07h2+14hx+3hhf'(x)=limh→07h2+14hx+3hh
Step 7.1.8
Factor hh out of 7h2+14hx+3h7h2+14hx+3h.
Step 7.1.8.1
Factor hh out of 7h27h2.
f′(x)=limh→0h(7h)+14hx+3hhf'(x)=limh→0h(7h)+14hx+3hh
Step 7.1.8.2
Factor hh out of 14hx14hx.
f′(x)=limh→0h(7h)+h(14x)+3hhf'(x)=limh→0h(7h)+h(14x)+3hh
Step 7.1.8.3
Factor hh out of 3h3h.
f′(x)=limh→0h(7h)+h(14x)+h⋅3hf'(x)=limh→0h(7h)+h(14x)+h⋅3h
Step 7.1.8.4
Factor hh out of h(7h)+h(14x)h(7h)+h(14x).
f′(x)=limh→0h(7h+14x)+h⋅3hf'(x)=limh→0h(7h+14x)+h⋅3h
Step 7.1.8.5
Factor hh out of h(7h+14x)+h⋅3h(7h+14x)+h⋅3.
f′(x)=limh→0h(7h+14x+3)hf'(x)=limh→0h(7h+14x+3)h
f′(x)=limh→0h(7h+14x+3)hf'(x)=limh→0h(7h+14x+3)h
f′(x)=limh→0h(7h+14x+3)hf'(x)=limh→0h(7h+14x+3)h
Step 7.2
Reduce the expression by cancelling the common factors.
Step 7.2.1
Cancel the common factor of hh.
Step 7.2.1.1
Cancel the common factor.
f′(x)=limh→0h(7h+14x+3)h
Step 7.2.1.2
Divide 7h+14x+3 by 1.
f′(x)=limh→07h+14x+3
f′(x)=limh→07h+14x+3
Step 7.2.2
Reorder 7h and 14x.
f′(x)=limh→014x+7h+3
f′(x)=limh→014x+7h+3
f′(x)=limh→014x+7h+3
Step 8
Step 8.1
Split the limit using the Sum of Limits Rule on the limit as h approaches 0.
limh→014x+limh→07h+limh→03
Step 8.2
Evaluate the limit of 14x which is constant as h approaches 0.
14x+limh→07h+limh→03
Step 8.3
Move the term 7 outside of the limit because it is constant with respect to h.
14x+7limh→0h+limh→03
Step 8.4
Evaluate the limit of 3 which is constant as h approaches 0.
14x+7limh→0h+3
14x+7limh→0h+3
Step 9
Evaluate the limit of h by plugging in 0 for h.
14x+7⋅0+3
Step 10
Step 10.1
Multiply 7 by 0.
14x+0+3
Step 10.2
Add 14x and 0.
14x+3
14x+3
Step 11
Step 11.1
Multiply 14 by 1.
m=14+3
Step 11.2
Add 14 and 3.
m=17
m=17
Step 12
The slope is m=17 and the point is (1,10).
m=17,(1,10)
Step 13
Step 13.1
Use the formula for the equation of a line to find b.
y=mx+b
Step 13.2
Substitute the value of m into the equation.
y=(17)⋅x+b
Step 13.3
Substitute the value of x into the equation.
y=(17)⋅(1)+b
Step 13.4
Substitute the value of y into the equation.
10=(17)⋅(1)+b
Step 13.5
Find the value of b.
Step 13.5.1
Rewrite the equation as (17)⋅(1)+b=10.
(17)⋅(1)+b=10
Step 13.5.2
Multiply 17 by 1.
17+b=10
Step 13.5.3
Move all terms not containing b to the right side of the equation.
Step 13.5.3.1
Subtract 17 from both sides of the equation.
b=10-17
Step 13.5.3.2
Subtract 17 from 10.
b=-7
b=-7
b=-7
b=-7
Step 14
Now that the values of m (slope) and b (y-intercept) are known, substitute them into y=mx+b to find the equation of the line.
y=17x-7
Step 15