Answer:
yes
Step-by-step explanation:
the FIRST derivative of a function tells us the slope of a tangent line to the curve at any point. if is positive, then the curve must be increasing. If is negative, then the curve must be decreasing.
the SECOND derivative gives us the slope of the slope function (in other words how fast the slope of the original function changes, and if it is accelerating up - positive - or if it is avengers down - negative).
so, the first derivative would be fully sufficient to get the answer of if the slope of the function at that point is positive or negative.
but because it is only a "if" condition and not a "if and only if" condition, the statement is still true.
there are enough cases, where the slope is positive, but the second derivative is not > 0 (usually = 0).
but if even the second derivative is positive, then, yes, the slope of the original function must be positive too.
Is there an image i don’t understand what expressions you’re asking for?
First Consider two points ~ (0 , -3) and (-1 , 0)
Now, use the given formula to calculate slope :
Therefore, the slope of the graph is -3
The graphs of f(x) and g(x) are transformed function from the function y = x^2
The set of inequalities do not have a solution
<h3>How to modify the graphs</h3>
From the graph, we have:
and
To derive y < x^2 - 3, we simply change the equality sign in the function f(x) to less than.
To derive y > x^2 + 2, we perform the following transformation on the function g(x)
- Shift the function g(x) down by 2 units
- Reflect across the x-axis
- Shift the function g(x) down by 3 units
- Change the equality sign in the function g(x) to greater than
<h3>How to identify the solution set</h3>
The inequalities of the graphs become
y < x^2 - 3 and y > x^2 + 2
From the graph of the above inequalities (see attachment), we can see that the curves of the inequalities do not intersect.
Hence, the set of inequalities do not have a solution
Read more about inequalities at:
brainly.com/question/25275758
Answer:44
Step-by-step explanation: