Graph question (2 Viewers)
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(i) Increasing.
The sign of the derivative shows when the curve is increasing (>0) or decreasing (<0) because derivative is the gradient function. i.e. input a point and it outputs the gradient at the point to the curve or gradient of tangent.
Increasing: gradient always positive so curve is always rising.
Decreasing: gradient always negative so curve is always falling.
As you can see in the graph, f' is always positive, so gradient is always positive from x = 0 to x = infinity. Looks like it has a horizontal asymptote at y = 0 hence will never go below 0. So f is increasing.
(ii) Concave down.
The sign of the second derivative shows when curve is concave up or concave down. It is also the gradient function of f'.
Concave up: like a frown. Rise up then stop at stationary point, then fall down.
Concave down: like a smile. Fall down then stop at stationary point, then rise up.
From graph, f' always has negative gradient so we say f'' < 0 for all real x. Hence f'' is concave down.
The sign of the derivative shows when the curve is increasing (>0) or decreasing (<0) because derivative is the gradient function. i.e. input a point and it outputs the gradient at the point to the curve or gradient of tangent.
Increasing: gradient always positive so curve is always rising.
Decreasing: gradient always negative so curve is always falling.
As you can see in the graph, f' is always positive, so gradient is always positive from x = 0 to x = infinity. Looks like it has a horizontal asymptote at y = 0 hence will never go below 0. So f is increasing.
(ii) Concave down.
The sign of the second derivative shows when curve is concave up or concave down. It is also the gradient function of f'.
Concave up: like a frown. Rise up then stop at stationary point, then fall down.
Concave down: like a smile. Fall down then stop at stationary point, then rise up.
From graph, f' always has negative gradient so we say f'' < 0 for all real x. Hence f'' is concave down.
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keepLooking
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i) By inspecting the f'(x) graph, you know that in the domain, f'(x) is always > 0, hence you know f(x) is increasing throughout its domain.
EDIT: psyc1011 has a better answer above ;D
EDIT: psyc1011 has a better answer above ;D
BlueGas
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This is the part the confuses me. I've seen these types of graphs before and I know the the graph will never touch the x-axis, even though the curve is going down it'll never touch the x-axis so how is f(x) increasing?
The graph displayed is f'(x), the gradient function. Assuming that it has a horizontal asymptote at y = 0, then it never touches the x-axis. That means f'(x) > 0, so always positive! Doesnt matter if the f' graph is swirling/jumping crazily as long as it is > 0 for all real x.
Positive gradient means it will always rise up and never fall down. i.e. increasing
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By inspecting the curve, the gradient function (f') itself is decreasing (starts off with high gradient then lowers down near 0), so we can say f is increasing at a decreasing rate.
Example values,
x = 0.01, y' = 100
x = 0.2, y' = 5
x = 5, y' = 0.2
x = 100, y' = 0.01
Since y' is clearly decreasing and always positive, then y' is increasing at a decreasing rate
Example values,
x = 0.01, y' = 100
x = 0.2, y' = 5
x = 5, y' = 0.2
x = 100, y' = 0.01
Since y' is clearly decreasing and always positive, then y' is increasing at a decreasing rate
BlueGas
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Yeah that's a much better way to word it, because the graph is actually increasing but the numbers get smaller, that's what confused me, so how about the second part, if it's basically a "bowl" shape it's always concave down and if it's a "hill" shape then it's concave up? Is it always like this?
BlueGas
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How about if a question was referring to y = x^3, when f'(x) = 3x^2, so looking at the graph 3x^2 it seems that it's increasing so the curve must be increasing too? But when looking at x^3 it doesn't seem like it's increasing.
Yes, always. But with respect to f(x), not f'(x).
If f' is increasing, then f is increasing is a logical error because that is the incorrect relation. f' can be negative, hence f is not always increasing.
Example, y = x^2 and y' = 2x. y' is increasing on all x but y isn't increasing on all x.
As for f(x) = x^3,
f' is not increasing. It is decreasing from -inf to 0 then increases from 0 to +inf.
f' is positive for all real x hence f is increasing for all real x.
Look at y = x^3 and check it's gradient at all points. Just draw tangents at all points on the graph. Do you see that all the tangents have a positive gradient? Hence, increasing for all x.
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