For what values of x does the graph of F9x) = 2x^3 - 3x^2 + 12x + 87 have a horizontal tangent?
this is the same as asking where F'(x) = 0
Since F'(x) = 6x^2-6x+12
where would that be?
6x4=24
2x6=12+12=24
so...2?
No, it's almost a trick question.
6x^2-6x+12 is never zero (has no real roots)
So, F(x) never has a horizontal tangent.
Visit wolframalpha.com and type in
2x^3 - 3x^2 + 12x + 87
and it will show the graph and various properties of the function. You will be able to see clearly that there is no horizontal tangent.
To find the values of x for which the graph of F(x) has a horizontal tangent, we need to find the points where the derivative of F(x) is equal to zero.
First, let's find the derivative of F(x) by applying the power rule to each term:
F'(x) = d/dx (2x^3) - d/dx (3x^2) + d/dx (12x) + d/dx (87)
= 6x^2 - 6x + 12
Next, we set the derivative equal to zero and solve for x:
6x^2 - 6x + 12 = 0
To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation:
x = (-(-6) ± √((-6)^2 - 4(6)(12))) / (2(6))
= (6 ± √(36 - 288)) / 12
= (6 ± √(-252)) / 12
Since we have a square root of a negative number, the quadratic equation has no real solutions. Therefore, there are no values of x for which the graph of F(x) has a horizontal tangent.