Find values for a and b such that f(x)= (ax+b)/(x+1)^2 has a critical number at x=-2.
You need to remember that the critical value is a zero for the first derivative of function, hence you need to differentiate the function with respect to x using quotient rule such that:
`f'(x) = ((ax+b)'*(x+1)^2 - (ax+b)*((x+1)^2)')/((x+1)^4)`
`f'(x) = ((a*(x+1)^2 - 2(ax+b)*(x+1)))/((x+1)^4)`
You need to factor out x+1 such that:
`f'(x) = ((x+1)(a*(x+1) - 2(ax+b)))/((x+1)^4)`
Reducing by x+1 yields:
`f'(x) = (ax + a - 2ax + 2b)/((x+1)^3)`
The problem provides the information that `x = -2` is a critical value to function f(x), thus `f'(-2) = 0 ` .
`f'(x) = (-ax + a + 2b)/((x+1)^3)`
`f'(-2) = 0 =gt f'(-2) = (2a + a + 2b)/((-2+1)^3)`
`3a + 2b = 0 =gt a = -2b/3`
Notice that the problem provides just one condition and you need more than one (at least two) since you need to find two unknowns, a and b.
Hence, evaluating a and b under given conditions yields `a = -2b/3.`