Question

If f(x)=10^x and 10^1.04 is 10.96, which is closest to f'(1)?
a)0.24
b)0.92
c)0.96
d)10.5
e)24

Answer 1

@Michele_Laino

Answer 2

we have to find the first derivative of 10^x

Answer 3

that is 10^x*ln10 right?

Answer 4

that's right!

Answer 5

so what is:
\[f'\left( 1 \right) = {10^1} \times \ln 10 = ...?\]

Answer 6

its 10ln10, I cannot use a calculator for this

Answer 7

ok!

Answer 8

I believe the question has something to do with local linearization but I'm not sure

Answer 9

\[f(x) \approx f'(1.04)(x-1.04)+f(1.04)\]

Answer 10

I don't know f'(1.04)

Answer 11

you can find f' from f
by differentiating f

Answer 12

that gives me 10.96ln10

Answer 13

we can linearize the function 10^x around the point x=1.04

Answer 14

ok

Answer 15

the only bad thing is they tell us what they want us to use for 10^(1.04)
but they don't tell us what they want us to use for ln(10)

Answer 16

thats my main problem

Answer 17

and you aren't allowed to use a calculator to figure out ln(10), right?

Answer 18

yes!

Answer 19

unfortunately no

Answer 20

do you know wath is log_10(e) ?

Answer 21

nope

Answer 22

\[\ln(10)=y \\ \text{ equivalently this is } e^y=10 \\ \text{ we know } e \approx 2.7 \\ 2.7^y=10 \\ \text{ this equation would give a } y \text{ close \to this other equation's y } \\3^y=10\]
y is like 2.something

Answer 23

let's use the approximate that y is 2.1
and see what we get at the end

Answer 24

we will use ln(10) is approximately 2.1

Answer 25

ok sounds good

Answer 26

\[f(x) \approx f'(1.04)(x-1.04)+f(1.04) \\ f'(x) \approx 10.96 \cdot 2.1(x-1.04)+10.96 \]

Answer 27

and then plug 1 into that right?

Answer 28

yah

Answer 29

ok thank you

Answer 30

another way is :
since we have:
e < 10 <e^(5/2)
then we can write:
1<ln(10)<5/2
and finally
10.96 < f '(1) <25

Answer 31

10 < f '(1) <25

Answer 32

I think something is wrong above

Answer 33

not with what @Michele_Laino said

Answer 34

I'm so dumb

Answer 35

I accidentally changed f(x) to f'(x)

Answer 36

\[f(x) \approx f'(1.04)(x-1.04)+f(1.04) \\ f(x) \approx 10.96 \cdot 2.1(x-1.04)+10.96\]

this is what was suppose to say

but we were looking for f'(1) not f(1)

Answer 37

oh right

Answer 38

lol we should have just found f'
given f

Answer 39

\[f'(x)=\ln(10) \cdot 10^x \\ f'(1)=\ln(10) \cdot 10^{1}\]

Answer 40

I honestly don't know why they gave you 10^(1.04)

Answer 41

\[f'(1)=2.1 \cdot 10\]

Answer 42

which is closer to that 23 choice

Answer 43

lol i guess it was easier than i thought

Answer 44

brute force seems to work

Answer 45

maybe it was a trick

Answer 46

like I bet my students will try to use that liberalization method if I say this

Answer 47

you should try it on them and see what happens XD

Answer 48

you know what

Answer 49

actually we still could use that lineazation method

Answer 50

oops one sec

Answer 51

\[f(x) \approx f'(a)(x-a)+f(a) \\ f(x) \approx f'(1)(x-1)+f(1) \\ \text{ we want \to solve for } f'(1) \\ f(x)-f(1) \approx f'(1)(x-1) \\ \frac{f(x)-f(1)}{x-1} \approx f'(1)\]

Answer 52

and we were given f(x) for when x=1.04

Answer 53

\[f'(1) \approx \frac{f(x)-f(1)}{x-1} \text{ @ x=1.04 } \\ f'(1) \approx \frac{10^{1.04}-10^{1}}{1.04-1} \approx \frac{10.96-10}{1.04-1}\]

Answer 54

how sick is that! :)

Answer 55

ah thats a lot better

Answer 56

very satisfying

Answer 57

I am actually more satisfied with that too

Answer 58

i think thats what the question makers were going for

Answer 59

yep yep me too

Answer 60

it wasn't a trick after all

Answer 61

ok well thanks a ton for helping!

Answer 62

np