SOLUTION: The function $f(x)$ is defined only on domain $[-1,2]$, and is defined on this domain by the formula f(x) = 2x^2 - 8x + 1. What is the range of $f(x)$? Express your answer as an

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: The function $f(x)$ is defined only on domain $[-1,2]$, and is defined on this domain by the formula f(x) = 2x^2 - 8x + 1. What is the range of $f(x)$? Express your answer as an       Log On


   

Question 1204243: The function $f(x)$ is defined only on domain $[-1,2]$, and is defined on this domain by the formula
f(x) = 2x^2 - 8x + 1.
What is the range of $f(x)$? Express your answer as an interval or as a union of intervals
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

range of f%28x%29+=+2x%5E2+-+8x+%2B+1 if restricted to +-1%3C=x%3C=2, or domain [-1,2]
f%28-1%29+=2+%28-1%29%5E2+-+8+%28-1%29+%2B+1=11
f%282%29+=2+%282%29%5E2+-+8+%282%29+%2B+1=-7
so, range is:
{ f%28x%29 element R+: -7%3C=+f%28x%29+%3C=11 }
interval:
[-7,11]



Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


The final answer from the other tutor is correct, but only by chance. Her analysis of the problem is incomplete.

The given function is quadratic; its graph is an upward-opening parabola. You can't determine the range of a quadratic function over the prescribed domain simply by evaluating the function at the two endpoints of that domain. It is very possible that the minimum value of the function could occur between those endpoints.

Her analysis comes up with the correct range only because the minimum value of the function is at x=2, which is the right endpoint of the prescribed domain.

The correct method for finding the range of a quadratic function over a prescribed domain is to evaluate the function at THREE points -- the two endpoints of the domain and at the vertex of the parabola.