Answer to Question #145755 in Algebra for liam donohue

Question

Consider the function f(x)=−2x2+6x−9. f(x) is increasing on the interval (−∞,A] and decreasing on the interval [A,∞) where A is the critical number.
Find A

At x=A, does f(x) have a local min, a local max, or neither? You should be aware that there are two different tests that may work. Either use the Second Derivative test and look at concavity at A, or use the First Derivative test and examine the sign table.
Type in your answer as LMIN, LMAX, or NEITHER.

Answer 1

Ok, the given function is quadratic. The sign of main coeffient of the function is negative. So it is directed to the downward and it can have max point. There are two ways that we can find that point. The first is taking derivetive, the second is finding x₀ and y₀.

1-way)

f(x)=-2x²+6x-9;

f'(x)=-4x+6; ----> -4x+6=0; x₀=1.5. So A=1.5;

f(1.5)=-4.5;

We check the sign of the derivative of the function before and after x_0;

f'(0)=6, f'(2)=-2;

The sign of the first derivetive of the function before A is positive and after A is negative. That's why

in the interval (-∞;A] function is increasing and in the interval [A;∞) it is decreasing.

f''(x)=-4; the second derivative is negative. So in the interval (-∞;∞) function is concave;

The function has no LMIN point, and it has LMAX. LMAX is (1.5;-4.5);

2- way) The asked symbol A is equal to x₀. So the second way is easy one:

General form of the quadratic function is f(x)= ax²+bx+c.

We have f(x)=−2x²+6x−9. Therefore, a=-2, b=6, c=-9

x₀=-b/2a, x₀= -6/2*(-2)=-6/-4=1.5.

A=x₀=1.5;