Question

in the picture

Answer 1

@douglaswinslowcooper This one I actually dont get at all. I know you have to find the tangent line and fill in the chart but thats about it

Answer 2

Sorry if it seems like im asking for alot but im studying for finals sooo yah

Answer 3

For the f(x) row just plug in square root of the x it asks and round it to four decimals.

For the second row T(x)
[sqrt(2) - f(x) you put in for the x] / [2 - x it asks]

Answer 4

ok so do the sqrt of x for the values at the top of the chart

Answer 5

f(x) = sqrt(x)
The derivative, f'(x) = 1/(2sqrt(x))
at x = 2 the slope is 1/(2sqrt(2))
So, the equation of the tangent line is y = x/(2sqrt(2)) + b
Plug in the values you were given(2,sqrt(2)) and you have,
sqrt(2) = 2/(2sqrt(2)) + b
sqrt(2) = 1/sqrt(2) + b
b = sqrt(2) - 1/sqrt(2)
So, the equation of the tangent line is y = 1/(2sqrt(2)) + sqrt(2)-1/sqrt(2)

Answer 6

Now you just need to evaluate that equation for each value of x that you need and fill in the chart

Answer 7

oops, for the tangent line I meant y =x/(2sqrt(2)) + sqrt(2)-1/sqrt(2)

Answer 8

thisis for the T(x) correct?

Answer 9

Yes

Answer 10

ok so for the f(x) I just take the sqrt of the number at the top