# Finding the missing values of a function

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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}{(x-1)(x-4)} \\ &= \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... $$g(2) = \frac{a(2) + b}{(2-1)(2-4)} = \frac{2a + b}{-2}= \frac{2a + b}{-2} = -1$$ $$2a + b = 2\qquad (eqn. 1)$$ Now, let's the find derivative of $g$ and use the other piece of info we are given - that $g'(2) = 0$. $$g'(x) = \frac{a(x^2-5x +4)-(ax+b)(2x -5)}{(x^2 - 5x +4)^2}$$ Since we're just going to substitute in the point $g'(2) = 0$, we can use this messy form directly. $$g'(2) = \frac{a(2^2-5(2)+4)-(a(2)+b)(2(2)-5)}{(2^2-5(2)+4)^2} = \frac{a(-2)-(2a+b)(-1)}{(-2)^2} = \frac{-2a +2a + b}{4}$$ $$g'(2) = 0 \implies \frac{b}{4} = 0$$

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, 2020

If the graph of the function $g(x) = \frac{ax+b}{(x-1)(x-4)}$ has a horizontal tangent at $(2,-1)$, determine the values of $a$ and $b$.
The 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?