Question

Over which interval of the domain is function h decreasing?

Answer 1

Your answer is D.

Explanation
It is the only one that makes sense

Answer 2

The interval of the domain of the considered function for which the function is decreasing is none. The function is strictly increasing: Option D: The function is increasing only.

When do we say that a function is decreasing?

Suppose that the function is y = f(x)

If for an interval I, when x increases meanwhile staying in the interval I, the function's output decreases (or can stay same), then we say that the function is decreasing in the interval I.

This interval I was of values that x can take for this function, which is the domain of the considered function, so we say:

y = f(x) is decreasing function in an interval I if:

$f(x+\delta) \leq f(x) \: \forall x \in I : x+\delta \in I$

If we have:

$f(x+\delta) < f(x) \: \forall x \in I : x+\delta \in I$, then the function is called strictly decreasing in the interval I.

Similarly, there increasing( $f(x+\delta) \geq f(x) \: \forall x \in I : x+\delta \in I$ ) and strictly increasing($f(x+\delta) > f(x) \: \forall x \in I : x+\delta \in I$ ) functions.

For this case, the function is:

$h(x) = \left \{ {{2^x, x < 1} \atop \: \atop {\sqrt{x+3}, x \geq 1}} \right.$

The function is differentiable in either side of 1.

Sign of rate of function at a point tells whether its increasing or decreasing.

• For x < 1:

$h'(x)= 2^x ln(2)$
We know that:

$2^x > 0 \: \rm \forlall x \in \mathbb R$

And $ln(2) \approx 0.69 > 0$

Thus, $h'(x)= 2^x ln(2) > 0 \: \forall x \in \mathbb R$

So, rate is positive, the function is strictly increasing for this case.

• For x > 1:

$h'(x) = \dfrac{1}{2\sqrt{(x+3)}}$

For all x > 1, $\sqrt{x+3} > 0$, and so as h'(x)

So rate is positive and therefore strictly increasing for this case too.

• For x  = 1:

$h(1+\delta) - h(1)= \sqrt{1+\delta + 3} - \sqrt{1+3} = \sqrt{4+\delta} - \sqrt{4} > 0 \forall \delta > 0$

So rate of the considered function is positive,so function is strictly increasing all over the domain.

Thus, the interval of the domain of the considered function for which the function is decreasing is none. The function is strictly increasing: Option D: The function is increasing only.

Learn more about increasing functions here:
https://brainly.com/question/14330051

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