Which interval contains a local minimum for the graphedfunction?
1 answer:
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I dont know of you wanted the graph too so here
everything is in the pictures here, also sorry if I am wrong
Nota: am folosit Google Translate așa că sper sensul nu se pierde în traducere<span>
Sunt presupunând că t, g, și c sunt constante
</span>simplifica
f(x)=cg²t²x²
amintiți-vă că derivitive de af (x), în cazul în care a este o constantă este un lucru ori mai mare decât derivitve lui f (x)
f(x)=cg²t²(x²)
<span>Amintiti-va: derivitive lui
este 
</span>
<span>
</span>
<span>f'(x)=cg²t²(2x¹)
f'(x)=cg²t²(2x)
f'(x)=2cg²t²x</span>
Sunt presupunând că t, g, și c sunt constante
</span>simplifica
f(x)=cg²t²x²
amintiți-vă că derivitive de af (x), în cazul în care a este o constantă este un lucru ori mai mare decât derivitve lui f (x)
f(x)=cg²t²(x²)
<span>Amintiti-va: derivitive lui
<span>
</span>
<span>f'(x)=cg²t²(2x¹)
f'(x)=cg²t²(2x)
f'(x)=2cg²t²x</span>
1. Consider the function 
.
It is a linear function, so to draw its graph 2 points are enough:
f(4)=(3/4)*4-2=3-2=1
f(0)=(3-4)*0-2=-2
so 2 points on this line are (0, -2) and (4, 1). Check the picture 1:
2. The problem considers the inequality y>f(x)
a. the slope of the line y=ax+b is the coefficient of x, so a. In our case a=3/4, not -2
b. The graph of any linear inequality is a shaded region, not a line or a shaded line.
c. Either the part below the line, or the part above the line will be shaded since we are talking about an inequality.
Check (4, -3), picture 2, which is clearly below the line.
now here y=-3, and f(4)=(3/4)(4)-2=3-2=1, so in this region y<f(x)=(3/4)x-2
So clearly the region where the inequality holds, so the region to color, is the part above the line.
d. The line itself is not dashed, only if we have an "equal to or smaller than" or "equal to or larger than" inequality.
e. (0, 0) is in the region above the line so YES, it is a solution.
We can also check it like this: y=0, f(0)=(3/4)0-2=-2 so y>f(0)
f. The y-intercept is always f(0). In this case f(0)=-2
It is a linear function, so to draw its graph 2 points are enough:
f(4)=(3/4)*4-2=3-2=1
f(0)=(3-4)*0-2=-2
so 2 points on this line are (0, -2) and (4, 1). Check the picture 1:
2. The problem considers the inequality y>f(x)
a. the slope of the line y=ax+b is the coefficient of x, so a. In our case a=3/4, not -2
b. The graph of any linear inequality is a shaded region, not a line or a shaded line.
c. Either the part below the line, or the part above the line will be shaded since we are talking about an inequality.
Check (4, -3), picture 2, which is clearly below the line.
now here y=-3, and f(4)=(3/4)(4)-2=3-2=1, so in this region y<f(x)=(3/4)x-2
So clearly the region where the inequality holds, so the region to color, is the part above the line.
d. The line itself is not dashed, only if we have an "equal to or smaller than" or "equal to or larger than" inequality.
e. (0, 0) is in the region above the line so YES, it is a solution.
We can also check it like this: y=0, f(0)=(3/4)0-2=-2 so y>f(0)
f. The y-intercept is always f(0). In this case f(0)=-2