Question

If the tangent line to y = f(x) at (4, 2) passes through the point (0, 1), find f(4) and f '(4).
f(4)=2 , f'(4)=?

Answer 1

so you can find the tangent line you are given two points and the slope
use point-slope form of a line

Answer 2

Y1-Y0=m(x1-xo)?

Answer 3

actually you only need the slope

Answer 4

yeah

Answer 5

since that is what it asked for

Answer 6

you know the slope formula right?

Answer 7

\[m=\frac{y_1-y_2}{x_1-x_2}\]

Answer 8

m=y1-y0/x1-xo

Answer 9

is the other way around of how you put it

Answer 10

1 and 0's
1 and 2's
whatever

Answer 11

do you know how to use the formula?

Answer 12

is 1/4

Answer 13

let's see top is 2-1
and bottom is 4-0
looks great

Answer 14

yes

Answer 15

do I use y=mx+b?

Answer 16

b=2 if I do that with (4,2)

Answer 17

you don't need to
it just asked to find f'(4) and f(4)
and f'(4) is the slope of the tangent line at (4,2) to the curve y=f(x)

Answer 18

unless you want to find the tangent line ?

Answer 19

I do

Answer 20

it is just:
\[y-f(a)=f'(a)(x-a) \text{ is the tangent line \to the curve } y=f(x) \text{ at } (a,f(a))\]

Answer 21

you are given a is 4 here

Answer 22

\[y-f(4)=f'(4)(x-4)\]

Answer 23

now you found f(4) and f'(4)

Answer 24

right

Answer 25

thanks

Answer 26

np

Answer 27

problem the answer is not 2.

Answer 28

I already tried that

Answer 29

f(4) should be 2 it is given by (4,2)
f'(4) should be 1/4 that is the slope of the tangent line at (4,2) to the curve y=f(x)

Answer 30

oh yeah then I was right about the 1/4. in the question I put the f(4)

Answer 31

perfect thank you

Answer 32

yeah you already found f(4) earlier
we were finding f'(4) and we used the slope formula because that is what the derivative means

Answer 33

Find
f '(a).
f(x) =

(1 − 2x)^1/2

Answer 34

so do you know chain rule?

Answer 35

yeah, saw this last year tho

Answer 36

I thought it was 1/2(1-2x)^-1/2 * (2)

Answer 37

\[h(x)=f(g(x)) \\ h'(x)=g'(x) \cdot f'(g(x)) \\ \\ \text{ if } f(x)=x^n \\ \text{ then we have } \\ h(x)=(g(x))^n \\ h'(x)=g'(x) \cdot n(g(x))^{n-1}\]

Answer 38

well one complaint

Answer 39

what is the derivative of 1-2x ?

Answer 40

you put 2 but it should actually be...

Answer 41

ooooh

Answer 42

thanks

Answer 43

but you can also simplify a bit

Answer 44

hahaha hate when that happens

Answer 45

yes i did before

Answer 46

in my homework

Answer 47

ok then you got it now :)

Answer 48

do you have time for more?

Answer 49

I had problems with this one too

Answer 50

maybe one more then I have to go check on something

Answer 51

(a) Find the slope m of the tangent to the curve
y = 7/

x
at the point where x = a > 0.

Answer 52

is that just y=7/x?

Answer 53

(b) Find equations of the tangent lines at the points (1, 7) and

(4, 7/2)


.

Answer 54

\[y=\frac{7}{x}=7 \cdot \frac{1}{x} =7 x^{-1} \text{ by law of exponents } \\ \text{ now differentiate } y=7 x^{-1} \text{ using power rule }\]

Answer 55

no x^1/2 sorry

Answer 56

oh

Answer 57

it didn't appear when i copy pasted it

Answer 58

I had -7/2x^3/2

Answer 59

but is not that

Answer 60

\[\text{ find the slope of the tangent line \to the curve } y=7x^\frac{1}{2} \text{ at } x=a>0\]?

Answer 61

or is that x^(1/2) in the denominator?

Answer 62

ok part A i got it

Answer 63

I had to change x=a

Answer 64

now for B I don't now what I did wrong

Answer 65

ok so it was:
\[y=\frac{7}{x^\frac{1}{2}}=7x^\frac{-1}{2}\]

Answer 66

I got that

Answer 67

but I have part B wrong

Answer 68

ok the tangent line at (1,7)
is
\[y-f(a)=f'(a)(x-a) \\ \text{ here you have } a=1 \\ y-f(1)=f'(1)(x-1) \\ \text{ you already found } f'(x) \text{ \above } \\ \text{ you got that } f'(x)=\frac{-7}{2x^{\frac{3}{2}}} \text{ I think but I don't really know anymore } \\ \text{ \because I'm still trying \to figure out what } f(x) \text{ was :p } \\ \text{ so this becomes } \\ y-7=\frac{-7}{2(1)^\frac{3}{2}}(x-1)\]

Answer 69

and you still have to do this same process for the other point

Answer 70

perfect

Answer 71

is not that apparently