Intermediate Value Theorem

• IVT

• EVT

• MVT

Selected values of a continuous function f are given in the table. What is the fewest possible number of zeros of f in the interval [0,5]?

• Zero, because $f\left(x\right)$ is not equal to 0 for any of the values in the table

• One, because f is continuous on the interval [0,5] and $f\left(0\right)>0>f\left(5\right)$

• Two, because the values of $f\left(x\right)$ in the table change from positive to negative twice

• Three, because f is continuous on the interval [0,5] and $f\left(0\right)>0>f\left(1\right),\ f\left(1\right)<0<f\left(3\right),\ f\left(3\right)>0>f\left(5\right).$

• More than three, because $f\left(x\right)$ could be equal to zero at the values not specified by the table

Tags

CCSS.HSA.APR.B.3

 $f\left(x\right)=3x-2\sin x$  

The Intermediate Value Theorem guarantees that that there is a solution on the closed interval [10,12] to which of the below equations?

•  $f\left(x\right)=0$  

•  $f\left(x\right)=17.372$  

•  $f\left(x\right)=32.421$  

•  $f\left(x\right)=67.205$  

Tags

CCSS.HSA.REI.B.3

CCSS.HSA.REI.D.11

Discontinuous functions possess the Intermediate Value Theorem property.

(check the appropriate box)

• Always

• Sometimes

• Never

Does  $f\left(x\right)=2+x-x^2$  holds with IVT given  $\left[0,3\right];\ w=1$  ?

• IVT holds

• IVT does not hold

• no sufficient information

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.REI.B.4

TRUE OR FALSE?  If f is continuous on [-1,1], f(-1)=4 and f(1)= -2, then there is a zero between -1 and 1.

• TRUE
• FALSE
• CANNOT BE DETERMINED

Let f be a function that is continuous on the interval [1, 10]. If f(2) = -5 , f(5) = 5 and f(9) = -5, which must be true?

I. f(x) has a least 2 zeros

II. For some c, 2 < c < 9, f(c) = 3

III. The graph of f(x) has a horizontal tangent

• I only

• I and II

• I, II, and III

• II and III

Tags

CCSS.HSA.APR.B.3

For

 $f\left(x\right)=x^2+4x+3$ , find the values(s) of "c" guaranteed by the IVT such that  $f\left(c\right)=2$   

•  $-3\pm\sqrt{2}$  

•  $4\pm\sqrt{5}$  

•  $-2\pm\sqrt{3}$  

•  $-4\pm\sqrt{5}$  

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.B.4

CCSS.HSA.REI.D.11