Question

please help me work through this homework problem, I got a similar problem wrong so I would like some guidance :)

Answer 1

The given function is:

`f(x)=2x^3-42x^2+240x-11`

It is required to find the x values for which the graph has a horizontal tangent line.

Recall that horizontal lines have a slope of 0.

Recall also that the derivative of a function at a point is the same as the slope of the tangent line at that point.

Hence, to find the x values at which the graph of the function has horizontal tangent lines, find the derivative of the function and equate it to zero.

Find the derivative of the function:

`\begin{gathered} f^(\prime)(x)=3\cdot2x^(3-1)-2\cdot42x^(2-1)+240x^(1-1)+0 \\ \Rightarrow f^(\prime)(x)=6x^2-84x+240 \end{gathered}`

Equate the derivative to zero and solve the resulting equation for x:

`\begin{gathered} 6x^2-84x+240=0 \\ \text{ Divide through by 2:} \\ 3x^2-42x+120=0 \\ \text{ Rewrite }42x\text{ as }-30x-12x\text{ and factorize the left-hand side:} \\ 3x^2-30x-12x+120=0 \\ 3x(x-10)-12(x-10)=0 \\ \Rightarrow(3x-12)(x-10)=0 \\ \Rightarrow3x-12=0\text{ or }x-10=0 \\ \Rightarrow3x=12\text{ or }x=10 \\ \Rightarrow x=4\text{ or }x=10 \end{gathered}`

Hence, the x-values are x=4 and x=10.

The required x values are 4 and 10.