Problem

Use the evidence that \(f(x)~=~x~+~5\) and \(g(x)~=~2x~+~1\) cross at \(x~=~4\) to compare the functions on \(0~\le~x~\le~4\). State where one is greater and where they are equal.

Big Picture

What this problem is really about

A larger slope tells which line is catching up, but not which one is already higher. We’ll subtract the functions so their comparison becomes a sign question, then use the known crossing to split the interval at exactly the right place. A quick endpoint check will confirm what happens before the meeting point and at the meeting point itself.

Answer choices

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Preview only

Since g increases faster and catches f at x = 4, g is greater for 0 <= x < 4 and then they are equal at x = 4.

Because the functions cross at x = 4, they must have been equal for all x from 0 to 4 instead of meeting only at that one x-value.

Since f starts higher at x = 0 and the lines do not cross until x = 4, f is greater for 0 <= x < 4 and they are equal at x = 4.

Since g has the larger slope, g is greater on 0 <= x <= 4, so it must stay above f all the way until they meet at x = 4.

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Use the evidence that \(f(x)~=~x~+~5\) and \(g(x)~=~2x~+~1\) cross at \(x~=~4\) to compare the functions on \(0~\le~x~\le~4\). State where one is greater and where they are equal.

What this problem is really about

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