For what values of x is the tangent line of the graph of
f(x)=2x^3−12x^2+19x+0
parallel to the line y=1x+0.7 ? Enter the x values in order, smallest first, to 4 places of accuracy:
the slope of the line is 1. So, you want the slope of the tangent to be 1.
f'(x) = 6x^2-24x+19 = 1
x = 1,3
check:
f(1) = 9
f(3) = 3
So the tangent lines are
y-9 = x-1, or y=x+8
y-3 = x-3, or y=x
the graphs are at
http://www.wolframalpha.com/input/?i=plot+y%3D2x%5E3%E2%88%9212x%5E2%2B19x,+y%3Dx%2B8,+y%3Dx
Why did the tangent line cross the road? To parallel park! Haha, just kidding. Let's find the values of x where the tangent line is parallel to y=1x+0.7.
First, let's find the derivative of f(x):
f'(x) = 6x^2 - 24x + 19
We know that when two lines are parallel, their slopes are equal. So, we need to find the x-values where f'(x) = 1.
6x^2 - 24x + 19 = 1
Simplifying the equation, we get:
6x^2 - 24x + 18 = 0
Dividing through by 6, we have:
x^2 - 4x + 3 = 0
Now we can factor this equation:
(x - 1)(x - 3) = 0
Setting each factor equal to zero, we get:
x - 1 = 0 or x - 3 = 0
Solving these equations, we find:
x = 1 or x = 3
So, the values of x where the tangent line is parallel to y=1x+0.7 are:
x = 1 and x = 3.
Remember, laughter is always a tangent to solving equations.
To find the values of x for which the tangent line of the graph of f(x) is parallel to the line y = x + 0.7, we need to find the derivative of f(x) and set it equal to the slope of the given line, which is 1.
Let's find the derivative of f(x):
f(x) = 2x^3 − 12x^2 + 19x + 0
f'(x) = 6x^2 − 24x + 19
Now set f'(x) equal to 1:
6x^2 − 24x + 19 = 1
Rearranging the equation:
6x^2 − 24x + 18 = 0
Dividing through by 6:
x^2 − 4x + 3 = 0
Factoring the quadratic equation:
(x − 1)(x − 3) = 0
Setting each factor equal to zero:
x - 1 = 0 or x - 3 = 0
Solving for x:
x = 1 or x = 3
So, the values of x for which the tangent line is parallel to y = x + 0.7 are 1 and 3.
To find the values of x for which the tangent line of the function f(x) is parallel to the line y = x + 0.7, we need to determine when the slopes of the two lines are equal.
First, let's find the slope of the tangent line for f(x). The slope of a function at a given point can be found by taking the derivative of the function. In this case, we'll take the derivative of f(x) with respect to x:
f(x) = 2x^3 − 12x^2 + 19x + 0
f'(x) = 6x^2 − 24x + 19
Now, we need to find the x-values where the slope of the tangent line is equal to 1, since the line y = x + 0.7 has a slope of 1.
Set f'(x) equal to 1 and solve for x:
6x^2 − 24x + 19 = 1
6x^2 − 24x + 18 = 0
x^2 − 4x + 3 = 0
To solve this quadratic equation, we can factor it:
(x − 1)(x − 3) = 0
This gives us two solutions: x = 1 and x = 3.
So, the values of x for which the tangent line of f(x) is parallel to y = x + 0.7 are x = 1 and x = 3.