How do you find the points where the graph of the function # f(x) = x^3 + 9x^2 + x + 19# has horizontal tangents?

Question

Samantha Adkison · Accepted Answer

Horizontal tangents occur when the tangents have a slope of #0#.

Differentiating, by the power rule: #f'(x) = 3x^2 + 18x + 1# The slope of the tangent is given by evaluting #f(a)#, where a is the point #x = a# where the tangent intersects the function. We can therefore say #f(x) =0# and solve the equation. #0 = 3x^2 + 18x + 1# #-1 = 3(x^2 + 6x + 9 - 9)# #-1 = 3(x + 3)^2 -27# #26/3 = (x + 3)^2# #+-sqrt(26/3) - 3 = x# Therefore, the tangent line is horizontal at the points #x = sqrt(26/3) - 3# and #x = -sqrt(26/3) - 3#. Hopefully this helps!

Elijah Abram · Answer

undefined To find the points where the graph of the function f(x) = x^3 + 9x^2 + x + 19 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of f(x) by applying the power rule: f'(x) = 3x^2 + 18x + 1.

Next, we set f'(x) equal to zero and solve for x: 3x^2 + 18x + 1 = 0.

Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = 18, and c = 1, we can calculate the values of x.

After solving the quadratic equation, we find two values for x: x ≈ -5.27 and x ≈ -0.73.

Therefore, the points where the graph of the function has horizontal tangents are approximately (-5.27, f(-5.27)) and (-0.73, f(-0.73)).