<span>The graph you plotted is the graph of f ' (x) and NOT f(x) itself. </span>
Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get
f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24
So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5
<span>f ' (x) = (x-4)(6-2x)
</span><span>f ' (3.5) = (3.5-4)(6-2*3.5)
</span>f ' (3.5) = 0.5
The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.
So the final answer is C) 3
If PQ is 14, each segment, PR and RQ will be 7.
14 = PR + 7.
Answer:
r=10
Step-by-step explanation:
Take 4 away from 5 to isolate your variable. Now you have r/10= 1. figure out the unknown value by placing ten on top which is the same as 1.
Here, we are required to determine how many hours your friend will drive in order to catch you.
(a)<em> Your friend will have to drive 7 and a half hours inorder to catch </em><em>you.</em>
<em>(</em><em>b)</em><em> </em><em>You </em><em>both </em><em>will </em><em>be </em><em>6</em><em>7</em><em>5</em><em> </em><em>miles</em> <em>away </em><em>from </em><em>Ellensburg</em><em> </em><em>at </em><em>that </em><em>time.</em>
If you leave at 1 pm; At 2:30pm;
- That is; 1 and a half hours after leaving; you must have covered a distance, d = 75 × 1.5
- d = 112.5miles.
Therefore, your position; S after 2:30pm is given by;
S(a) = 75t + 112.5 miles from Ellensburg.
For your friend; travelling at 90miles/hr;
- His position is given as; S(b) = 90 × t
(a) For your friend to catch you, you both must be in the same position;
75t + 112.5 = 90t
90t -75t = 112.5
t = 112.5/15
t = 7.5hours
(b) To determine how far you both are from Ellensburg; we can either evaluate:
S(b) = 90t or S(a) = 75t + 112.5
Therefore, By evaluating S(b) = 90t.
S(b) = 90 × 7.5
S(b) = 675miles from Ellensburg.
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