Finding the missing values of a function
I'm going to rewrite the denominator so that when we take the derivative, we don't have to use both the quotient rule and the multiplication rule. $$\begin{align}g(x) &= \frac{ax+b}{(x1)(x4)} \\ &= \frac{ax+b}{x^2  5x +4} \end{align}$$ Since we're given that the point $(2,1)$ is on the graph of $g$, substituting this in to the equation gives us one equation involving only the unknowns, a and b. So, let's do that... $$g(2) = \frac{a(2) + b}{(21)(24)} = \frac{2a + b}{2}= \frac{2a + b}{2} = 1$$ $$2a + b = 2\qquad (eqn. 1)$$ Now, let's the find derivative of $g$ and use the other piece of info we are given  that $g'(2) = 0$. $$g'(x) = \frac{a(x^25x +4)(ax+b)(2x 5)}{(x^2  5x +4)^2}$$ Since we're just going to substitute in the point $g'(2) = 0$, we can use this messy form directly. $$g'(2) = \frac{a(2^25(2)+4)(a(2)+b)(2(2)5)}{(2^25(2)+4)^2} = \frac{a(2)(2a+b)(1)}{(2)^2} = \frac{2a +2a + b}{4}$$ $$ g'(2) = 0 \implies \frac{b}{4} = 0$$
So $b = 0$ and substituting this result back into eqn. 1 we find that $a=1$.
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user481710
Updated on March 19, 2020Comments

user481710 over 3 years
If the graph of the function $g(x) = \frac{ax+b}{(x1)(x4)}$ has a horizontal tangent at $(2,1)$, determine the values of $a$ and $b$.
I am unsure on how to go about solving this question. I found the first derivative but am unsure of what to do next

Paul Z over 9 yearsThe first derivative was a great start. Now you have two equations: $g(2) = 1$ and $g'(2) = 0$; expand the definitions of $g$ and $g'$ to get two simultaneous equations in $a$ and $b$. Can you finish it from here?
