Need help in finding the slope of the tangent: f(x) = √16-x, where y = 5

[dungas]

The question is this: f(x) = √16-x, where y = 5. The answer is supposed to be -1/10.

What I've done:
lim = f(x + h) - f(x)/h
h->0

lim = √16 - (x + h) - (√16 - x)/h
h->0

I did radical conjugate after this and came up with this:


lim = 16 - x - h - 16 + x/h(√16 -x - h + √16 - x)
h->0

16 and -16 becomes 0 and so does -x and x. This is what I was left with.

lim = - h /h(√16 -x - h + √16 - x)
h->0

The -h and the h in the denominator cancel out and this is what is left.

lim = -1 /(√16 -x - h + √16 - x)
h->0

Then h goes to 0 (h -> 0) and this is what's left.


= -1 /2(√16 -x)


The answer is supposed to be -1/10. I don't know how to get this answer.


I would really appreciate it if you guys guided me to on how to get the correct answer.

 

[Deleted member 4993]

The question is this: f(x) = √16-x, where y = 5. The answer is supposed to be -1/10.

What I've done:
lim = f(x + h) - f(x)/h
h->0

lim = √16 - (x + h) - (√16 - x)/h
h->0

I did radical conjugate after this and came up with this:


lim = 16 - x - h - 16 + x/h(√16 -x - h + √16 - x)
h->0

16 and -16 becomes 0 and so does -x and x. This is what I was left with.

lim = - h /h(√16 -x - h + √16 - x)
h->0

The -h and the h in the denominator cancel out and this is what is left.

lim = -1 /(√16 -x - h + √16 - x)
h->0

Then h goes to 0 (h -> 0) and this is what's left.


= -1 /2(√16 -x)

You have missed many grouping symbols. If you were my student you would have lost 50% of credit by this time. Your answer looks like

\(\displaystyle - \dfrac{1}{2\sqrt{16-x}}\)

which should be written in ASCII characters as:

- 1/[2√(16-x)]

Now use the fact that:

√(16-x) = y

and continue.....

The answer is supposed to be -1/10. I don't know how to get this answer.


I would really appreciate it if you guys guided me to on how to get the correct answer.

.

 

[dungas]

Can you expand on that, I have no clue what you mean.

 

[Deleted member 4993]

Can you expand on that, I have no clue what you mean.

For which part of my reply - you need further elaboration?

 

[dungas]

For which part of my reply - you need further elaboration?

"Now use the fact that:

√(16-x) = y"

 

[pka]

"Now use the fact that: √(16-x) = y"

\(\displaystyle \begin{align*}\frac{\sqrt{16-x-h}-\sqrt{16-x}}{h}&=\frac{h}{h(\sqrt{16-x-h}+\sqrt{16-x})} \\&=\frac{1}{(\sqrt{16-x-h}+\sqrt{16-x})} \end{align*}\)

 

[Deleted member 4993]

"Now use the fact that:

√(16-x) = y"

- 1/[2√(16-x)] = - 1/[2*y] = - 1/10