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.

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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

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