exam 1 2009, question 10

[bar0029]

can anyone explain this please from last years paper?

The gradient of the function is positive but decreasing for x ≥ 8 . The gradient of the tangent at x = 8 is positive and
constant. The approximation is determined by the y-value on the tangent which is greater than the actual value on the
function for x>8.

[costa94]

yeah ok so graph the function
then graph a tangent to the function at the point where x equals 8

the value is taken as the y-coordinate of the tangent at the point just above 8 (whatever it was can't rmemeber now) which is higher up than the y-coordinate of the function at that actual point. therefore the y-coord or approximation u get is actually higher than what its meant to be.

[vea]

We were talking about it in here just before

http://vcenotes.com/forum/index.php/topic,32341.0.html

[thushan]

OK. Consider the function f(x) = x^(1/3), which is a cubic flipped onto the side.

We tried to work out f(8.06) by assuming the gradient of the function is CONSTANT over the interval [8, 8.06], in other words y is increasing at a constant rate over interval [8, 8.06], and we know that f( = 2. However, the gradient of the function is DECREASING, so y is increasing at (on average) a lower rate over this interval. So the approximate value (determined by assuming gradient is constant) is higher than the real value (as the gradient is in reality decreasing).