can any one answer me this mamatmatical questions with all necessary steps?

Question

Find the critical number of g(x) = 2x4+16x3-6
2) What is the equation of the tangent to the graph of f(x) = x2+3x-1 at x= 1
3) The slope of the tangent line of function f(x) = lnx at x= -1 is ____

1) f(x) = xcube-3xsquare+16, then

1.1The interval on w/h f strictly increasing is ______
1.2 The interval on w/h f strictly decreasing is ______
1.3 The local maximum of f at x = _______________
1.4 The local mimum of f at x = _______________

Answer 1

1) g(x) = 2x^4 + 16x^3 - 6

To find the critical numbers, we need the solution to the equation g'(x) = 0. Therefore, we solve for g'(x) and then make it 0.

g'(x) = 8x^3 + 48x^2. Setting g'(x) to 0,
0 = 8x^3 + 48x^2. Factoring 8x^2 out, we get
0 = 8x^2 (x + 6)

Therefore, 8x^2 = 0, x + 6 = 0. This implies
x^2 = 0, x = -6.
So x = {0, -6}, and these are our critical numbers.

2) f(x) = x^2 + 3x - 1, equation of tangent line at x = 1.

Note that if x = 1, then f(1) = 1 + 3 - 1 = 3. So we want the equation of the tangent line at (1, 3).

The slope of the tangent line is found by solving for f'(1). So we solve for f'(x).

f'(x) = 2x + 3.
Therefore, m = f'(1) = 2(1) + 3.
m = 5

We now use the slope formula, with (1, 3) and (x, y) as (x1, y1) and (x2, y2), and with the slope being m = 5.

(y2 - y1) / (x2 - x1) = m
(y - 3) / (x - 1) = 5

Multiplying both sides by (x - 1) gives us
y - 3 = 5(x - 1)
y - 3 = 5x - 5

y = 5x - 2 is the equation of the tangent line.

3) f(x) = ln(x)

Normally, we would find f'(-1) to obtain the slope. However, this is a trick question, because f(x) = ln(x) is not defined if x = -1 (you cannot take the log of a negative number). Therefore, it doesn't exist.

Watch what would have happened if we thought it would have existed:

f'(x) = 1/x, therefore f'(1) = m = 1/(-1) = -1
We would have had an answer when there wouldn't have been one.

4) f(x) = x^3 - 3x^2 + 16

To find the intervals of increase/decrease, as well as local extreme, we find the derivative and the critical numbers.

f'(x) = 3x^2 - 6x
0 = 3x(x - 2)

Therefore, x = {0, 2}

To determine intervals of increase/decrease, we must test a single point around those critical numbers. That is, we want to test
1) a value less than 0,
2) a value between 0 and 2
3) a value greater than 2

We want to test these for positivity/negativity, and we want to test them in f'(x) = 3x^2 - 6x, or f'(x) = 3x (x - 2).

Test -1: f'(-1) = 3(-1) (-1 - 2) = 3(-1)(-3) = 9, which is positive.
Therefore, f is increasing on (-infinity, 0]

Test 1: f'(1) = 3(1) (1 - 2) = 3(-1) = -3, which is negative.
Therefore, f is decreasing on [0, 2]

Test 100000: f'(100000) = 3(100000)(100000 - 2) = DEFINITELY POSITIVE. Therefore, f is increasing on [2, infinity)

The interval on which f is strictly increasing is
(-infinity, 0] U [2, infinity)

The interval on which f is strictly decreasing is
[0, 2]

To find the local max, we need to determine the point at which we switch from increasing to decreasing. This occurs at x = 0. Therefore, we solve for f(0), and get f(0) = 0^3 - 3(0)^2 + 16,
f(0) = 16. So we have a local max at (0, 16).

To find the local min, we determine where we switch from decreasing to increasing. This occurs at x = 2. Solving for f(2), we get f(2) = 2^3 - 3(2)^2 + 16 = 8 - 12 + 16 = 12.
So we have a local min at (2, 12),

Answer 2

You need a calculus tutor, not just quick Yahoo Answers.

I take g(x) to be 2x^4 + 16x^3 - 4, and I guess you know how to differentiate and get
g'(x) =8x^3 + 48x^2

Are critical numbers the values at which the derivative is zero? So you need to solve
8x^3 + 48x^2 = 0.

That is 8x^2(x + 6) = 0. I'll assume you can do that.

For 2), find f'(x). Sub x = 1 in f(x), and the value you get is y(1).

The equation of the tangent is y - y(1) = m(x -1), where m is the value of f'(1), i.e. sub x = 1 in the expression you got for f'(x).

3) In this case f'(x) = 1/x, sub x = -1 in this to find the slope of the tangent line.

For 1), again find f'(x), then

1.1 Solve f'(x) = 0, getting two values of x. f will be increasing either between those values, or else on the left of the smaller one and the right of the larger.

To find out which it is, sub in f'(x) any value of x between the values you found. Is it + or -? My guess is it's -, which means f is decreasing, so it's increasing before the lower value and after the higher value, so your answer might be something like
x < 0 or x > 2,
or if you use set notation:
{x: x < 0}U{x: x > 2}

1.2 You already found this one in doing 1.1. (Or, if you found it's increasing between those values, then it's decreasing between them)

1.3 You already found, in doing 1.1, the two values at which f'(x) = 0. At one of them, f is increasing before it and decreasing after , so it's a local maximum. at the other, it's decreasing before and then increasing after, so it's local minimum. So you've done 1.4 as well.

Congratulations!

If any of this isn't clear, email [e-mail]

Answer 3

take derivative of g(x). set that equal to 0.
The solutions are 0,0,-48. that is your critical #

2)take derivative of f(x). replace x with 1. the slope is 5. find f(x) for x=1. it is 4. so the answer is the line through the point (1,4), with slope of 5

3) undefined i think

1) this is the same as the first question, find citical numbers, plot (critical number, corresponding f(x)) points , connect them with a line, and answer the questions