CALCULUS HELP

Question

 

x       −1.5       −1       −0.5       0       0.5       1       1.5

f (x)       −1       −4       −6       −7       −6       −4     −1
f '(x)       −7      −5       −3        0         3          5         7

 

Let f be a differentiable function such that f ''(x) > 0 for [–1.5, 1.5]. The table above shows values of f(x) and f '(x) for selected values of x on the closed interval [–1.5, 1.5].

 

^That is the chart and the information that comes with it^

 

 

1. At what value of x does the graph of f(x) reach a relative minimum? Give one reason based on
the values of f(x) AND one reason based on the values of
f '(x).

 

2. Using the values of either f(x) or f '(x) , explain why the values in the table confirm the assertion f ''(x) > 0.

 

3. Write an equation of the tangent line to the graph of f when x = –1 and use the equation of the tangent line to approximate the value of f(–1.1).

 

4. Is the approximation in part c for f(–1.1) an under or over approximation of the actual value? Justify your answer.

 

 

I am very new to the site, so I do not know if I am able to ask this many questions at once, but please answer them, as they are very important.

 

 

Answer 1

 

 

1.  from the f(x) values we see the f is concave-up function: it has minimum.

     also, we see f'(0)=0 meaning f(0) is a relative extremum, namely MINIMUM, because

     f(0)=-7 and it's the smallest value compared to closest points in table ...

 

2.  one sees f'(x) is an increasing functions; that is, f''(x)>0

 

3.  tangent line at x=-1 has a slope of f'(-1)=-5. write the equation of line having slope -5 and crossing point

     (-1,-4):

 

     y+4=-5(x+1) you'll end up with y=-5x-9 [double check this ...] ,

     and f(-1.1)=5.5-9=-3.5

 

4. average change of f in interval [-1,-1.5] is -4+1=-3 units. this will average to change [-1,-1.1]

    to 3*0.1/0.5=0.6 units. we would expect something like f(-1.1)~-4.0+0.6=-3.4 , whereas we        approximated to -3.5 . looks as if the approximation is a bit underestimated ...

 

check the arithmetic in parts 3. and 4.

 

 

Good luck:)