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3 years ago

Write the equation of the function g(x) if g(x) = f(x+2) +4 and f(x) = x^3 -7

1 answer:

4 0

F(x) = x³ - 7
f(x + 2) = (x + 2)³ - 7
f(x + 2) = (x + 2)(x + 2)(x + 2) - 7
f(x + 2) = (x(x + 2) + 2(x + 2))(x + 2) - 7
f(x + 2) = (x(x) + x(2) + 2(x) + 2(2))(x + 2) - 7
f(x + 2) = (x² + 2x + 2x + 4)(x + 2) - 7
f(x + 2) = (x² + 4x + 4)(x + 2) - 7
f(x + 2) = (x²(x + 2) + 4x(x + 2) + 4(x + 2)) - 7
f(x + 2) = (x²(x) + x²(2) + 4x(x) + 4x(2) + 4(x) + 4(2)) - 7
f(x + 2) = (x³ + 2x² + 4x² + 8x + 4x + 8) - 7
f(x + 2) = (x³ + 6x² + 12x + 8) - 7
f(x + 2) = x³ + 6x² + 12x + 8 - 7
f(x + 2) = x³ + 6x² + 12x + 1

g(x) = f(x + 2) + 4
g(x) = (x³ + 6x² + 12x + 1) + 4
g(x) = x³ + 6x² + 12x + 1 + 4
g(x) = x³ + 6x² + 12x + 5

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3 years ago

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sukhopar [10]

Answer:

her first mistake was on line 3.

line 1: 80 + [2(3 1/2 + 1 1/2)]

line 2: 80 + [2(5)]

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Step-by-step explanation:

on line 3 she added 80 + 2, but she should have multiplied 2 by 5. the correct steps would be:

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4 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

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3 years ago

Each batch of ms.crawley's cookies makes 14 cookies. if sandy needs 8 dozen cookies for her bake s how many batches does she hav

erica [24]

7 batches. There is 12 cookies in a dozen so multiply 8×12 to get 96 cookies. 96 Is the total cookies you need to make. So if you divide 96 by the 14 to see how many batches you must make. you will get a number of 6.85... so you must round up to 7.

4 0

3 years ago