How do you find the points where the graph of the function #f(x) = x^3 - 4x^2 - 7x + 8# has horizontal tangents and what is the equation?

Question

Cameron Alexander · Accepted Answer

See below.

The tangent line is horizontal when the slope of the tangent line is #0#. The slope of the tangent line is #0# when the derivative is #0#. So to find the #x#-coordinates at which the tangent line is horizontal, we need to solve #f'(x) = 0#. For this function, #f(x) = x^3-4x^2-7x+8#, we get #f'(x) = 3x^2-8x-7 = 0# I assume a calculus student knows the quadratic formula or completing the square, so solve the equation to get #x = (4+-sqrt37)/3#. There are two places where the tangent line is horizontal. At #x = (4-sqrt37)/3#, we get #y = f((4-sqrt37)/3) = ((4-sqrt37)/3)^3-4((4-sqrt37)/3)^2-7((4-sqrt37)/3)+8# # = (-164+74sqrt37)/27# The tangent line is horizontal. The equation of a horizontal line through the point #(h,k)# is #y = k#, so the tangent line to the graph of #f(x) = x^3-4x^2-7x+8# at #x = (4-sqrt37)/3# is # y = (-164+74sqrt37)/27# At the other point we have #x = (4+sqrt37)/3#. So, #y = f((4+sqrt37)/3) = ((4+sqrt37)/3)^3-4((4+sqrt37)/3)^2-7((4+sqrt37)/3)+8# # = (-164-74sqrt37)/27# And the equation of that tangent line is # y = (-164-74sqrt37)/27#

Samantha Adkison · Answer

undefined To find the points where the graph of the function f(x) = x^3 - 4x^2 - 7x + 8 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 - 8x - 7.

Next, we set f'(x) equal to zero and solve for x: 3x^2 - 8x - 7 = 0.

Using the quadratic formula, x = (-(-8) ± √((-8)^2 - 4(3)(-7))) / (2(3)).

Simplifying this equation, we get x = (8 ± √(64 + 84)) / 6.

Further simplifying, x = (8 ± √148) / 6.

Finally, we can simplify the equation to x = (4 ± √37) / 3.

Therefore, the points where the graph of the function has horizontal tangents are (4 + √37) / 3 and (4 - √37) / 3.