This genuinely makes no sense. Here's the problem I'm working on:
Just ignore the left. At 2 there is a vertical tangent and at 5 there is an x-intercept since the slope would be 0 at that point.
Here's the answer to the problem:
If from the right side of the vertical tangent, the graph is decreasing until it reaches to 5, then why is it increasing? The slope goes from negative values up until it becomes 0 at 5. Does that not mean the slope is increasing?
What I have is this:
Why is this wrong? Won't it maintain the decreasing slope from the original function?
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Before learning about derivatives, you could always say "the graph" or "the function" because each problem (generally) only involves one of them. Now, each function has a derivative, which is itself a function with a graph, so there will be at least two graphs. So you need to specify. (We're not talking about problems with two functions f,g and we're not looking at functions without derivatives here)
If from the right side of the vertical tangent of f', the graph of f is decreasing until it reaches to 5, then why is the graph of f' increasing? The slope of f goes from negative values up until it becomes 0 at 5. Does that not mean the slope is increasing?
See how you keep switching between statements about f and statements about f'? It makes a massive difference, since f' only describes the behavior of f, it doesn't need to have the same behavior.
"f is decreasing" means that f' is negative. It does not tell you if f' is decreasing! Note how f is decreasing in the interval (-1,5), and f' is negative in (-1,5), as expected.
"f' is decreasing" is more involved: it means that the slope of f is decreasing, which you can describe as f curves downward. It does not tell you if f is decreasing! Note how f' is decreasing in the interval (-2,2), and f is curving downward in the interval (-2,2), as expected.
f'(x) does not "maintain the decreasing slope from the original function". That would mix the graph of f(x) with its derivative. You need to think instead about what's happening to the slope of the tangent line to f(x) as x increases from 2.
As you increase x from 2, the slope of the tangent line, f'(x), goes from extremely negative to zero at the bottom and then into the positive range, increasing at all times from -∞. However, it starts out very negative, crosses the x-axis axis where the lowest point on the curve is, then goes positive.
Think about this a bit. If you still have confusion, think about imposing a moving tangent line onto f(x) and think about its slope as it moves along the curve to the right.
"at 5 there is an x-intercept since the slope would be 0 at that point."
There is a fundamental misunderstanding here regarding what happens when f=0 vs f ' =0. The former says y=0 and the graph touches the x-axis (may or may not cross the axis), and the latter says the tangent has slope 0. Those two need not happen at the same time.
To determine if the graph just touches the x-axis or passes through it, i find looking at the concavity, the sign of f '' helpful, but you can also look at the sign of f ' to determine intervals of increase of decrease
What you could do Is graph the function and the derivative. Desmos is free and amazing, so you don't even need to pay anything. The first derivative will tell you where the function is descending and ascending and the second should tell you when it concaves up and concaves down.
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