Question

B. Sketch the graph of the following functions and then find the absolute extreme values of each of the given interval.

1.f(x)=3-2/5x;[-4,0]

2.f(x)=x-3/4+x;[-3,1]

C. Let f(x)=2x+1. Determine if the Intermediate value theorem applies to f on the closed interval [-3,4]for k=1.

Answer 1

Solution (a)

The sketch of the function $f(x)=3-\frac{2}{5x}$ for the interval $[-4,0]$ is shown below

The plot shows that the function $f(x)=3-\frac{3}{2x}$ is undefined for $x=0$ . Therefore, the point $x=0$ is the point of discontinuity.

From the plot, we can see that $x=-4$ is the point of absolute minima and the absolute minimum value is $f(-4)=3.1$

Absolute maximum value does not exist. No point of maxima.

Solution (b)

The sketch of the function $f(x)=x-\frac{3}{4+x}$ for the interval $[-3,1]$ is shown below

The plot shows that the function $f(x)=x-\frac{3}{4+x}$ is undefined for $x$ between the given interval $[-3,1]$ .

From the plot, we can see that

$x=-3$ is the point of absolute minima and the absolute minimum value is $f(-3)=-6$

$x=1$ is the point of absolute maxima, and the absolute maximum value is $f(1)=0.4$

Solution (c)

Given that $f(x)=2x+1$ for the interval $[-3,4]$

Here

$f(-3)=2(-3)+1\\f(-3)=-5$ and $f(4)=2(4)+1\\f(4)=9$

Since

$f(-3)<0$ and $f(4)>0$

Therefore, the intermediate value theorem applies to f on the closed interval $[-3,4]$