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Aleksandr [31]
3 years ago
5

Which is a valid prediction about the continuous function f(x)?

Mathematics
2 answers:
uranmaximum [27]3 years ago
6 0

f(x) ≥ 0 over the interval [–1, 1]. is the correct answer

grigory [225]3 years ago
5 0
<span> C f(x) ≥ 0 over the interval [–1, 1]. </span>
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If you wanted to shift the graph of y = 2x + 5 up, which equation could you use?
maxonik [38]

Answer:

y = 2x + 5+2 graph would shift by 2 unit.

Step-by-step explanation:

Given  : y = 2x + 5 .

To find :If you wanted to shift the graph up what would be equation .

Solution : We have given that y = 2x + 5 .

BY the transformation rule ; if f(x) +k then graph would shift up by k units .

Then to shift the graph of y = 2x + 5 up we need to add something in 5

Like   y = 2x + 5+2

Then graph would shift by 2 unit.

Therefore,   y = 2x + 5+2 graph would shift by 2 unit.

3 0
4 years ago
Read 2 more answers
It says 9x+45 (Geometry question)
faltersainse [42]

Answer: The answer is 4

3 0
3 years ago
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
What is the coefficient in the scientific notation 9.6 x 10-8? 9.6 8 10​
zzz [600]

Answer:

9.6

Step-by-step explanation:

5 0
2 years ago
What is the value of X
UkoKoshka [18]

Answer:

28b

Step-by-step explanation:

3 0
3 years ago
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