Find the derivative of f(x) = negative 9 divided by x at x = 6.

2 answers:

5 0

First, find f(x+h) = (-9/(x+h)) ... just substitue (x+h) for every x.

Then plug in f(x+h) and f(x) in the difference quotient formula.

Lim/h->0\ [f(x+h) - f(x)]/h

Or, [(-9/ (x+h)) - -9/x] /h
Or, [(-9/ (x+h)) + 9/x]/h
Or, [(-9x + 9x + 9h)/ x(x+h)] / h
Or, [9h / (x^2 + hx)]/h .............................................. add like terms
Or, ( 9h / (x^2 + hx)) * h/1 .................divide by 7 is equivalent to multiply by 1/7
Or, 9/ (x^2 + hx) .................................cancle out the h
Or, 9/ x^2 ........................................... plug in 0 for h

Now, the function asks for x=6, so we plug in 6 for every x
so, the limit = 9/ 6^2 = 9/36 = 1/4.

o-na [289]3 years ago

4 0

F(x) = -9/x f(6) = -9/6 f(6) = -3/2 f(6) = -1.5 The function of f(x) = -9 / x at x = 6 is equal to -3 / 2 or -1.5. Hope that helps!

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I need to know how to solve this equation for x

mel-nik [20]

I hope this helps. X is equal to 4 and -9

5 0

3 years ago

Can someone help me with this? I need to find the points of discontinuity/limits for each of these. I think one point is 4, but

Debora [2.8K]

The answers are shown in the attached image

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Explanation:

Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x

x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4

The x values 0 and 4 make the denominator zero

These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.

There are two vertical asymptotes

Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity

Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity

Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity

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Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity

Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side

So overall, as we approach x = 4, the y value is heading off to positive infinity

Again everything is summarized in the image attachment

Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.