<p>If f(2)=7 and f(12)=1 what is f(7)
A) 5.2
B) 5
C) 4.6
D) 4
E) 3.4</p>
<p>Please help. I searched for every way to solve this question but couldn't solve it. Really appreciate it</p>
<p>Need more info. f could be any function (even just a random mapping of points including (2,7) and (12,1)).</p>
<p>However, if f is linear, we note that 7 is halfway between 2 and 12, so (7, f(7)) must be the midpoint of (2,7) and (12,1). The average of 7 and 1 is 4, so f(7) = 4 (D).</p>
<p>The answer is 4 I think. Treat the number inside the parentheses as an input (x) value and the outside # as a y value. So u have 2 sets of points. Find thr slope. Now set that slope equal to the new set of points (7, y). Solve</p>
<p>Sent from my SCH-I535 using CC</p>
<p>I also recommend using the slope to get the answer</p>
<p>Yeah, finding the slope definitely works. In this case, 7 is halfway between 2 and 12, so the answer must be halfway between 1 and 7.</p>
<p>^ that is assuming it’s linear… the prob didn’t say that… need more info…
is this cb?!</p>
<p>it can’t be solved unless it’s linear otherwise he would provide us with an equation or something in the question</p>
<p>Use the mean value theorem…(f(a)-f(b))/a-b=f’(c). here f’(c) is your differentiated fn which is a constant. The fn has to be linear…I guess:)</p>
<p>but “he” didn’t say it’s linear…
Point is… it’s very vague…</p>
<p>quiverfox… translate bro…
didn’t get a thing ._."</p>
<p>Quiverfox, what the hell?</p>
<ol>
<li> No SAT question requires calculus.</li>
<li> The Mean Value Theorem says only that there must be some point c in the interval (a,b) for which f’(c) = [f(b)-f(a)]/(b-a), not that [f(b)-f(a)]/(b-a) must be the slope of the tangent at every point in the interval (a,b).</li>
</ol>
<p>Mitcho, a long time ago, rspence was correct: you can’t answer this question without knowing that f is a linear function. But if f is indeed linear, then you’ve been given two points on the line: (2,7) and (12,1). The long way to do this is to find the equation of the line through those points, then plug x=7 into that equation and find the resulting value of y. The shorter way to do it is to recognize that 7 is halfway from 2 to 12, and to realize that if f is linear, then f(7) will be halfway between f(2), which is 7, and f(12), which is 1. In other words, f(7) will be 4, because 4 is halfway between 7 and 1.</p>
<p>But all of this depends on the assumption that f is linear.</p>
<p>^ yeah that’s what i thought… u think any question wud come that vague?</p>
<p>Yeah I sorta realised…:P</p>
<ul>
<li>I recommend that everyone ignores this post.</li>
</ul>
<p>@quiver </p>
<p>The mean value theorem can’t be applied here. In order to use the MVT we would need to know that the function was continuous on [2,12] and differentiable on (2,12). The problem does not indicate this. Furthermore, the MVT only gives some value c between 2 and 12 for which the slope of the tangent line at c is equal to -3/5. There is no guarantee that c would be equal to 7.</p>