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2 years ago

If f(x)=x^3/2 then f'(4)=

1 answer:

5 0

First you have to find the derivative of the equation f(x)= $x^{3/2}$ . To do this, you can use the power rule ( $f(x)= x^{n}$ goes to $f'(x)=nx^{n-1}$ ). so the derivative would be f'(x)= $\frac{3}{2} x^{ \frac{3}{2}-1 }$ , which can be written as f'(x)= $\frac{ 3\sqrt{x}}{2}$ . then you plug in 4 for x, and get $\frac{3 \sqrt{4} }{2} =3$ , so f'(x)=3

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A car travels 15 kilometers west in 10 minutes. After reaching the destination, the car travels back to the starting point, agai

Ugo [173]

Your answer to your question is C

4 0

2 years ago

Read 2 more answers

Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to

nikklg [1K]

Answer:

Part 1)

See Below.

Part 2)

$\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)$

Step-by-step explanation:

Part 1)

The linear approximation L for a function f at the point x = a is given by:

$\displaystyle L \approx f'(a)(x-a) + f(a)$

We want to verify that the expression:

$1-36x$

Is the linear approximation for the function:

$\displaystyle f(x) = \frac{1}{(1+9x)^4}$

At x = 0.

So, find f'(x). We can use the chain rule:

$\displaystyle f'(x) = -4(1+9x)^{-4-1}\cdot (9)$

Simplify. Hence:

$\displaystyle f'(x) = -\frac{36}{(1+9x)^{5}}$

Then the slope of the linear approximation at x = 0 will be:

$\displaystyle f'(1) = -\frac{36}{(1+9(0))^5} = -36$

And the value of the function at x = 0 is:

$\displaystyle f(0) = \frac{1}{(1+9(0))^4} = 1$

Thus, the linear approximation will be:

$\displaystyle L = (-36)(x-(0)) + 1 = 1 - 36x$

Hence verified.

Part B)

We want to determine the values of x for which the linear approximation L is accurate to within 0.1.

In other words:

$\displaystyle \left| f(x) - L(x) \right | \leq 0.1$

By definition:

$\displaystyle -0.1\leq f(x) - L(x) \leq 0.1$

Therefore:

$\displaystyle -0.1 \leq \left(\frac{1}{(1+9x)^4} \right) - (1-36x) \leq 0.1$

We can solve this by using a graphing calculator. Please refer to the graph shown below.

We can see that the inequality is true (i.e. the graph is between y = 0.1 and y = -0.1) for x values between -0.179 and -0.178 as well as -0.010 and 0.012.

In interval notation:

$\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)$

4 0

2 years ago

A radar gun was used to record the speed of a swimmer (in meters per second) during selected times in the first 2 seconds of a r

Natalka [10]

Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]

the area of each trapezoid is (v(t1)+v(t2))/2 times width

for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8

2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24

3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75

4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7

add them all up
0.8+2.24+1.75+0.7=5.49

5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters

8 0

2 years ago

Choose the correct simplification of (3x3 + 2x2 − 5x) − (8x3 − 2x2)

zmey [24]

(3x³ + 2x² - 5x) - (8x³ - 2x²) =
3x³ + 2x² - 5x - 8x³ + 2x² =
-5x³ + 4x² - 5x

5 0

2 years ago

Read 2 more answers

Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the b

Aloiza [94]

Answer:

a.) 0.1028

b.) 0.6477

c.) 0.0388

d.) 3

e.) 2.55

Step-by-step explanation:

Forming a binomial Probability distribution

n = 20

Probability of success for Weld metal failure = 85%

Probability of success for base metal failure = 15%

We use the probabilit distribution formula of combination to solve the problem.

P(x=r) = nCr * p^r * q^n-r

a.) if exactly 5 are base metal failures, then p = 15 and our solution becomes:

P(x=5) = 20C5 * 0.15^5 * 0.85^15

P(x=5) = 0.1028

b.) probability that fewer than 4 are base metal failure= P(x=0) + P(x=1) + P(x=2) + P(x=3)

P(x=0) = 20C0 * 0.15^0 * 0.85^20 = 0.0388

P(x=1) = 20C1 * 0.15¹ * 0.85^19 = 0.1368

P(x=2) = 20C2 * 0.15² * 0.85^18 = 0.2293

P(x=3) = 20C3 * 0.15³ * 0.85^17 = 0.2428

Probability that fewer than 4 are base metal failures becomes: 0.038 + 0.1368 + 0.2293 + 0.2428 = 0.6477

c.) probability that none of them are results of base metal failure = P(x=0). As earlier calculated,

P(x=0) = 0.0388

d.) mean of base metal failures = np = 20*0.15 = 3

e.) standard deviation of base metal failures = √np(1-p)

=3 * (1 - 0.15) = 3 * 0.85

= 2.55

4 0

2 years ago