Incorrect 00:00 Evaluate the expression for x = 1.7. Enter your answer in the box. х2 + 3x

Mathematics

1 answer:

IrinaK [193]3 years ago

5 0

Answer:

7.99

Step-by-step explanation:

x^2 +3x

Let x=1.7

(1.7)^2 +3(1.7)

2.89+5.1

7.99

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Mr. Lopez wrote the equation 32g+8g-10g=150

Brrunno [24]

Answer: g=5

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Relative extrema of f(x)=(x+3)/(x-2)

Salsk061 [2.6K]

Answer:

$\displaystyle f(x) = \frac{x + 3}{x - 2}$ has no relative extrema when the domain is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Step-by-step explanation:

Assume that the domain of $\displaystyle f(x) = \frac{x + 3}{x - 2}$ is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivative of this function at $x$ .

Since this domain is an open interval, $x = a$ is a relative extremum of this function if and only if $f^{\prime}(a) = 0$ and $f^{\prime\prime}(a) \ne 0$ .

Hence, if it could be shown that $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ , one could conclude that it is impossible for $\displaystyle f(x) = \frac{x + 3}{x - 2}$ to have any relative extrema over this domain- regardless of the value of $f^{\prime\prime}(x)$ .

$\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}$ .

Apply the product rule and the power rule to find $f^{\prime}(x)$ .

$\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}$ .

In other words, $\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

Since the numerator of this fraction is a non-zero constant, $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ . (To be precise, $f^{\prime}(x) < 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace\!$ .)

Hence, regardless of the value of $f^{\prime\prime}(x)$ , the function $f(x)$ would have no relative extrema over the domain $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .