4. Let g be the function that satisfies g(0)=0

1 answer:

4 0

We have that for the Question "Find expressions for g(x) and g′′(x)." it can be said that expressions for g(x) and g′′(x) is

a) $g(x) = (-x^2,x$

b) $g''(x)= (-2,x$

From the question we are told

Let g be the function that satisfies g(0)=0 and whose <em>derivative </em>satisfies g′(x)=2|x|.

<h3>The Expressions for g(x)</h3>

Generally the equation for the g'(x) is mathematically given as

$g'(x) = (-2x,x$

Since

$g(x=\int{g'(x)})$

$g(x)=g'(x) = (-x^2,x$

Therefore

$g''(x)= (-2,x$

For more information on Expressions visit

brainly.com/question/19007362

You might be interested in

True or false (Picture provided)

Naya [18.7K]

<h2>Hello!</h2>

The answer is: True.

A counterexample is a way that we can prove that something is not true about a mathematical equation or expression, it's also considered as an exception to a rule.

So:

$sec^{2}x-1=\frac{cosx}{cscx}\\\\tg^{2}x=\frac{cosx}{\frac{1}{sinx}}\\\\\frac{1-cos2x}{1+cos2x}=cosxsinx$

Then, evaluating we have:

$\frac{1-cos(2*45)}{1+cos(2*45)}=cos(45)*sin(45)\\\\\frac{1-0}{1+0}=\frac{\sqrt{2} }{2}*\frac{\sqrt{2}}{2}\\\\1=\frac{(\sqrt{2})^{2} }{4}\\\\1=\frac{2}{4}\\\\1=\frac{1}{2}$