- c) is the derivative of a)
- f) is the derivative of d)
- b) is the derivitive of h)
and g) and e) are irrelevant
Assuming u know:When the derivative function = 0 at a point x, then y at x is a turning point (trough or a crest)
first one because the turning points of a), (ie when gradient is 0), at at the x-intersections of c). This is the same for second one.
How ever you might think that e) is the derivative of c):
-the thing is that y value e) goes from negative to positive in the first intersection (from left)
-and the first (from left) turning point is a crest, which goes from positive and negative gradient
- so e) is not the derivative of c)
If you integrate c) you get a power-4 graph, and we've used the only one
If you differentiate it, you get a straight line: which there is none
So we are left with b), g) and h):
Personally, Ive never seen a graph like h) before.
But this is what i think:
lets divide the graph curves into positive x (right) and negative x (left)
consider b) having a derivative
left: the gradient is positive
right: the gradient is positive
both other graphs have curves negative y, so b) doesnot have a derivative.
Consider g)
left:gradient is negative
right: gradient is positive
y values of h) are always negative and b) is positive on the left and negative on the right (opposite)
So g) also does not have a derivative
So h) does have one
gradient
-left: positive
-right:negative
-also when x approaches 0, the gradient is infinitely negative/positive
And these properties are shown in the curve b), and so... the third pair.
Now Ive said a bit tooo much