Need a little help

[jgarcia]

The question reads:

compute the derivative of the given function and find the slope of the line that is tangent to its graph for the specified value of the independent variable.

a) f(x)=2; c=13

Here is my work...

f'(x)=0 therefore m=0

then y-y1=m(x-x1)

y-2=0(x-13)
y=0(x-13+2)
y=0x-11
y=0-11
y=-11

So my answer is ... f'(x)=0; y=-11

HOWEVER, the answer is supposed to be ... f'(x)=0; y=2

My Ys are different. I don't know where I made my mistake!?

then there is this question:

b) f(x)=7-2x; c=5

Here is my work...

f'(x)=7 --> 0 and -2x --> -2

f'(x)=-2 therefore m=-2

f(x) = y = 7-2x

y=7-2(5)
y=7-10
y=-3

Then

y-y1=m(x-x1)
y-(-3)=-2(x-(5))
y+3=-2(x-5)
y=-2x-5-3
y=-2x-8

so my answer is f'(x)=-2; y=-2x-8

But the answer should be f'(x)=-2; y=-2x+7

Again, my Ys are different. I don't know where I made my mistake!?

 

[Quaid]

The question reads:

compute the derivative of the given function and find the slope of the line that is tangent to its graph for the specified value of the independent variable.

f(x) = 2; c=13

Here is my work...

f'(x)=0 therefore m=0

then y-y1=m(x-x1)

y-2=0(x-13)
y=0(x-13+2)

Here is your mistake. On the right-hand side, you added 2 inside the grouping symbols, instead of outside.

2 is not a part of the number x-13

y - 2 = 0(x - 13)

y = 0(x - 13) + 2

y = 0 + 2


Perhaps, next time, you can simplify the expression 0(x-13) first because zero times anything is zero.

y - 2 = 0(x - 13)

y - 2 = 0

Now add 2 to both sides!

Cheers

 

[Quaid]

then there is this question:

f(x)=7-2x; c=5

Here is my work...

f'(x)=7 --> 0 and -2x --> -2

f'(x) = -2

I don't understand what the line highlighted in red is about.

f'(x) is not 7

y+3=-2(x-5)
y=-2x -5 -3

Arithmetic mistake here; -2 times -5 is not -5.

Cheers

PS: In future threads, one exercise per thread, please

 

[jgarcia]

Thanks so much!

much appreciated