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4 years ago

Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).

Mathematics

1 answer:

Setler79 [48]4 years ago

3 0

Answer:

$\Huge \boxed{3a+4}$

Step-by-step explanation:

The function is given.

$f(x)=3x+1$

To find : $f(a+1)$

The input for the function f(x) is (a + 1).

Replace x with (a + 1).

$f(a+1)=3(a+1)+1$

Expand brackets.

$f(a+1)=3a+3+1$

Simplifying.

$f(a+1)=3a+4$

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Given that mean of quiz scores = 6.4 and standard deviation = 0.7

And we need to use Chebyshev's theorem to find the range in which 88.9% of data will reside.

Chebyshev's theorem states that "Specifically, no more than $\frac{1}{k^{2} }$ of the distribution's values can be more than k standard deviations away from the mean".

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4 years ago

Find the Derivative.<br> f(x) = -xcos3x

DENIUS [597]

Answer:

$\frac{d}{dx}(-xcos3x)= 3x \ sin(3x)- cos(3x)$

Step-by-step explanation:

Use the Product Rule for derivatives, which states that:

$\frac{d}{dx} =[f(x)g(x)]= f(x)g'(x)+f'(x)g(x)$

In the function we are given, $f(x)=-x \cos3x$ , we can break it up into two factors: -x is being multiplied by cos3x.

Now, we have the factors:

$-x \\ $cos 3x$

Before using the product rule, let's find the derivative of cos3x using the chain rule and the power rule.

$\frac{d}{dx}(cos3x)= \frac{d}{dx} (cos3x) \times \frac{d}{dx} (3x) \\ \frac{d}{dx}(cos3x)= (-sin3x) \times 3 \\ \frac{d}{dx} (cos3x) = -3sin(3x)$

Now let's apply the product rule to f(x) = -xcos3x.

$\frac{d}{dx}(-xcos3x)= (-x)(-3sin3x) + (-1)(cos3x)$

Simplify this equation.

$\frac{d}{dx}(-xcos3x)= (x)(3sin3x) - (cos3x)$

Multiply x and 3 together and remove the parentheses.

$\frac{d}{dx}(-xcos3x)= 3x \ sin(3x)- cos(3x)$

Therefore, this is the derivative of the function $f(x)=-xcos3x$ .

3 0

3 years ago

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