Find the values of x at which the curve has a horizontal tangent line.
f(x) = (x-3)^4/x^2 + 2x
i really need help on this one. i look in my textbook i can't find any examples to help me or guide me with this
Find the values of x at which the curve has a horizontal tangent line.
f(x) = (x-3)^4/x^2 + 2x
i really need help on this one. i look in my textbook i can't find any examples to help me or guide me with this
I would be stunned if your calculus book has no examples of this problem. You perhaps do not recognize what it is asking, but there are definitely examples.
Recall what it means to be tangent to a curve. And recall the behavior of the derivative at a point on the curve where the tangent is a horizontal line.
The problem is simply a matter of finding where the derivative behaves like a horizontal line. What value is that (for the derivative).
No, you need to use the quotient rule and chain rule.
I'll get you started:
$\displaystyle f(x) = \frac{(x-3)^4}{x^2} + 2x$
$\displaystyle f'(x) = \frac{x^24(x-3)^3 - (x-3)^4(2x)}{x^4} + 2 $
$\displaystyle f'(x) = \frac{4x^2(x-3)^3 - 2x(x-3)^4}{x^4} + 2 $
Let f'(x) = 0 and solve for any x value(s).
You should note that x is in the denominator, and the denominator of a fraction can never equal zero, therefore x=0 is a non-permissible value.
If you must do this algebraically, it's quite long, you would have something like this:
$\displaystyle 0 = \frac{4x^2(x-3)^3 - 2x(x-3)^4}{x^4} + 2$
$\displaystyle -2= \frac{4x^2(x-3)^3 - 2x(x-3)^4}{x^4}$
$\displaystyle -2x^4= {4x^2(x-3)^3 - 2x(x-3)^4}$
Then multiplying out everything, bringing it all over to one side of the equation and solving, it would be very difficult.
If you can use a calculator just punch in $\displaystyle f'(x) = \frac{4x^2(x-3)^3 - 2x(x-3)^4}{x^4} + 2$ and find wherever it intersects the x axis.
There's two values, x = -3.133 and x = 1.888.
Em Yeu Anh's solution is for the function
$\displaystyle f(x)=\frac{(x-3)^4}{x^2}+2x.$
If the function is
$\displaystyle f(x)=\frac{(x-3)^4}{x^2+2x},$
then using the Quotient Rule, which states that
$\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2},$
we obtain
$\displaystyle f'(x)=\frac{(x^2+2x)\cdot 4(x-3)^3
-(x-3)^4\cdot(2x+2)}{(x^2+2x)^2}.$
Here, we have used the fact that $\displaystyle \frac{d}{dx}(x-3)^4=4(x-3)^3$, which may be checked by hand or found by the Chain Rule. Now we find the values of $\displaystyle x$ for which $\displaystyle f'(x)=0$:
$\displaystyle \frac{(x^2+2x)\cdot 4(x-3)^3
-(x-3)^4\cdot(2x+2)}{(x^2+2x)^2}=0.$
We know that if $\displaystyle x=0$, $\displaystyle f'(x)$ is undefined and the tangent line cannot be horizontal. Excluding that case, we multiply both sides by $\displaystyle (x^2+2x)^2$ to obtain
$\displaystyle (x^2+2x)\cdot 4(x-3)^3
-(x-3)^4\cdot(2x+2)=0.$
We readily see one solution: $\displaystyle x=3$. We therefore know that one horizontal tangent line will be found at $\displaystyle (3,f(3))$. Assuming we are anywhere else, we can divide both sides by $\displaystyle (x-3)^3$, which gives
$\displaystyle \begin{aligned}
4(x^2+2x)-(x-3)(2x+2)&=0\\
(4x^2+8x)-(2x^2+2x-6x-6)&=0\\
2x^2+12x+6&=0\\
x^2+6x+3&=0.
\end{aligned}$
Because the final statement is equivalent to the first, solving this quadratic equation will give us the other values of $\displaystyle x$ at which the tangent line of $\displaystyle f(x)$ is horizontal.
Hope this helps!