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3 years ago

Which of the following is NOT correct regarding exponential function y = a*, a > 0,a CANNOT = 1

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

5 0

I would choose c because it makes the most sense rather then the rest

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You are given the function f(x)=1/x^2+10. f(x) is concave up whenever: A. x^2−10 is negative. B. 3x^2−10 is positive. C. 3x^2+10

lukranit [14]

Answer:

f(x) is concave up whenever:

B. 3x²−10 is positive

f(x) is concave down whenever:

A. 3x²−10 is negative

The points of inflection of f(x) are the same as:

B. the zeros of 3x²−10

Step-by-step explanation:

Given the function f(x) = 1 / (x²+10)

We can determine the concavity by finding the second derivative.

f"(x) > 0 ⇒ f(x) is concave up

f"(x) < 0 ⇒ f(x) is concave down

Then

f'(x) = (1 / (x²+10))' = -2x / (x²+10)²

⇒ f"(x) = -2*(10-3x²) / (x²+10)³

if f"(x) = 0 ⇒ -2*(10-3x²) = 0 ⇒ 3x²-10 = 0

f(x) is concave up whenever 3x²−10 > 0

f(x) is concave down whenever 3x²−10 < 0

The points of inflection of f(x) are the same as the zeros of 3x²-10

it means that 3x²-10 = 0

7 0

3 years ago

Please find the result !

Sliva [168]

Answer:

$\displaystyle - \frac{1}{2}$

Step-by-step explanation:

we would like to compute the following limit:

$\displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right)$

if we substitute 0 directly we would end up with:

$\displaystyle\frac{1}{0} - \frac{1}{0}$

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

now notice that after simplifying we ended up with a rational expression in that case to compute the limit we can consider using L'hopital rule which states that

$\rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right)$

thus apply L'hopital rule which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right)$

simplify which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right)$

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

$\displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } }$

simplify which yields:

$\displaystyle - \frac{1}{2}$

finally, we are done!

6 0

3 years ago

Read 2 more answers

There are 6800 students at a college, 3860 of them are female. What percent of students are female?

Alisiya [41]

Answer:56.76%

Step-by-step explanation:

3860 is 56.76% of 6800

8 0

3 years ago

How much will the taxi driver earn if he takes one passenger 4.8 miles in another passenger 7.3 miles explain your process

Tcecarenko [31]

Answer:

12.1x+t

Step-by-step explanation:

since we do not get a specific amount we must add 4.8 and 7.3 and substitute the amount the driver earns per mile with x and use t as a variable if they charge something like $5 just to get in before driving

6 0

2 years ago

The hours of daylight, y, in Utica in days x from January 1, 2013 can be modeled by the equation y = 2.89sin(0.0145x-1.40) + 10.

Pie

Answer:

9.3 hours

Step-by-step explanation:

Given

$y = 2.89\sin(0.0145x-1.40) + 10.99.$