I need help with the following question.
Find the largest positive value of x at which the curve:
$y = (2x + 7)^6 (x - 2)^5$
has a horizontal tangent line.
I need help with the following question.
Find the largest positive value of x at which the curve:
$y = (2x + 7)^6 (x - 2)^5$
has a horizontal tangent line.
To solve y'(x)=0 when $y(x)\ne0$, one can consider the derivative \dfrac{y'(x)}{y(x)} of the function $\log|y(x)|=6\log|2x+7|+5\log|x-2|$. The computations become much simpler since \frac{y'(x)}{y(x)}=\frac{6\cdot2}{2x+7}+\frac5{x-2}. Thus, y'(x)=0 as soon as the RHS is zero, that is, when $12(x-2)+5(2x+7)=0$, that is, when $x=-\frac12$. Complete the reasoning with the values where $y(x)=0$, that is, $x=-\frac72$ and $x=2$.
Hint: what slope corresponds to a horizontal tangent? How do you find the slope of a tangent line?
Hint: $\displaystyle{\frac{dy}{dx} = \left(12(2x+7)^5(x-2)^5+5(2x+7)^6(x-2)^4 \right) = 0}$ , in other words
$\displaystyle{\frac{dy}{dx} = (2x+7)^5(x-2)^4(22x+11) = 0}$ at what points?
$\displaystyle{\frac{dy}{dx} = 0}$ at $x=-\frac{7}{2}, x=2, x=-\frac{1}{2}$