Question 1190389: Let f(x)=6-x-x^2
Find the open intervals of f which is increasing(decreasing)
A. The relative maxima of f occur at x=
B. the relative minima of f occur at x=
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
If this is for a calculus class, then you would start off by finding the derivative
f(x) = 6 - x - x^2
f ' (x) = -1 - 2x
Use the power rule that says x^n has the derivative of nx^(n-1).
Then set this equal to zero and solve for x
f ' (x) = 0
-2x-1 = 0
-2x = 1
x = -1/2
Now we apply the first derivative test.
Let's plug something in for x that's smaller than -1/2
I'll use x = -1
f ' (x) = -2x-1
f ' (-1) = -2(-1)-1
f ' (-1) = 1
We get a positive value
f ' (x) is positive on the open interval (-infinity, -1)
aka the interval -infinity < x < -1
Therefore, f(x) itself is increasing on this open interval
Now try something to the right of -1/2. I'll use x = 0
f ' (x) = -2x-1
f ' (0) = -2(0)-1
f ' (0) = -1
We get a negative value, so f(x) is decreasing on the interval -1 < x < infinity which in interval notation is (-1, infinity)
To summarize so far:
f(x) increases when -infinity < x < -1
f(x) decreases when -1 < x < infinity
The change from increasing to decreasing points us to a local or relative maximum.
The local max occurs when x = -1/2
Plug this into the original f(x) function to find its paired y value
f(x) = 6 - x - x^2
f(-1/2) = 6 - (-1/2) - (-1/2)^2
f(-1/2) = 25/4
f(-1/2) = 6.25
The location of the relative max point is exactly (-1/2, 25/4) which in decimal form is (-0.5, 6.25) and that's also exact.
If this isn't for a calculus class, then I recommend using your calculators "maximum" feature/tool.
You can plot the original f(x) curve using a tool like Desmos and the local max point should show up.
Click on the point itself to have its coordinates pop up.
Even in calculus settings, such graphing tools are useful to check your answer.
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