<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
I got x=18
3(x-4)-2(3x+4)=4(3-x)+5x+4
3x-12-6x-8=12-4x+5x+4
3x-20=16+x
2x=36x
x=18
(9x^4-13x^3-x-7)+(7x^3-2x+1)=
9x^4-13x^3-x-7+7x^3-2x+1=
9x^4+(-13+7)x^3+(-1-2)x-7+1=
9x^4+(-6)x^3+(-3)x-6=
9x^4-6x^3-3x-6
Answer: Option <span>D.)9x^4−6x^3−3x−6</span>
it would = out to be 9 18/73
Area= length * height
21*9 = 189cm^2
