1) For both of these functios, decide whether the function is always decreasing, the function is always increasing, or the function is sometimes increasing and sometimes decreasing.
a) x^3-3x+1
b) y=1-3x-x^3
2) Find the coordinates of the point on this curve at which the gradient is zero and determine whether the point is a local maximum or local minimum, giving reasons:
y= x^2+3/square root of x
3) A circular pipe has outer diameter of 4cm and thickness t cm.
a) show that the area of cross-section, Acm^2, is given by A= pi(4t-t^2).
b) Find the rate of increase of A with respect to t when t=1/4 and when t=1/2, leaving pi in the answer.
Can someone help me with the above? It would be much appreciated.
1) For both of these functios, decide whether the function is always decreasing, the function is always increasing, or the function is sometimes increasing and sometimes decreasing.Originally Posted by Confuzzled?
a) x^3-3x+1
b) y=1-3x-x^3
In both cases find the derivative, for example in a):. The function is increasing on any intervals for which
and decreasing on intervals for which
.
2) Find the coordinates of the point on this curve at which the gradient is zero and determine whether the point is a local maximum or local minimum, giving reasons:
y= x^2+3/square root of x
For "gradient" I presume you mean "derivative?"
Again, we need the derivative:. Solve for when this is zero, and that gives you your possible x values. Now, to determine which x values give local minima or maxima you can 1) graph the original equation and take a look, or 2) use the second derivative test.
The second derivative test says that when x is a candidate for a local minimum or maximum thatfor a local minimum and
for a local maximum. (Note how the signs of y'' appear to be "backwards" from what you might expect.) By the way, if
then x is neither a local max or min, but something called an "inflection point." This is a point where the curvature goes from positive to negative.
-Dan
3) A circular pipe has outer diameter of 4cm and thickness t cm.
a) show that the area of cross-section, Acm^2, is given by A= pi(4t-t^2).
b) Find the rate of increase of A with respect to t when t=1/4 and when t=1/2, leaving pi in the answer.
I presume by cross-section we are talking about the area of the physical pipe, not the empty space inside it.
a) The area of the whole pipe (including the empty space at the center) is. (Recall that 4 cm is a diameter, not a radius!) The area of the empty space is
. So the cross-section is (dropping the unit for a moment)
.
b) The rate of change of A (increase or decrease) is, so take your derivative of A (with respect to t) and plug in your values.
-Dan
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