f<span>(x)</span>=<span>x^2</span>−<span>6
</span>Replace <span>f<span>(x)</span></span> with <span>yy</span>.
<span>y=<span>x^2</span>−<span>6
</span></span>Interchange the variables.
<span>x=<span>y2</span>−6
</span>Solve for <span>yy</span><span>.
</span>
Move <span><span>−6</span><span>-6</span></span> to the right side of the equation by subtracting <span><span>−6</span><span>-6</span></span> from both sides of the equation.<span><span><span>y2</span>=6+x</span><span><span>y2</span>=6+x</span></span>Take the <span><span>square</span><span>square</span></span> root of both sides of the <span><span>equation</span><span>equation</span></span> to eliminate the exponent on the left side.<span><span>y=±<span>√<span>6+x</span></span></span><span>y=±<span>6+x
</span></span></span>The complete solution is the result of both the positive and negative portions of the solution.
Tap for more steps...
<span>y=<span>√<span>6+x</span></span>,−<span>√<span>6+x</span></span></span>
Solve for y<span> and replace with </span><span><span>f^<span>−1</span></span><span>(x).
</span></span>
<span>Answer is f<span>−1</span></span><span>(x)</span>=<span>√<span>6+x</span></span>,−<span>√<span>6+<span>x</span></span></span>
Answer:
B
Step-by-step explanation:
This answer shows values of -2, and less than -2
Write an expression using multiplication and addition with a sum of 16:
2 x 6 +4
Well this is simple a calculator type problem...but if you are curious as the the algorithm used by simple calculators and such...
They use a Newtonian approximation until it surpasses the precision level of the calculator or computer program..
A newtonian approximation is an interative process that gets closer and closer to the actual answer to any mathematical problem...it is of the form:
x-(f(x)/(df/dx))
In a square root problem you wish to know:
x=√n where x is the root and n is the number
x^2=n
x^2-n=0
So f(x)=x^2-n and df/dx=2x so using the definition of the newton approximation you have:
x-((x^2-n)/(2x)) which simplifies further to:
(2x^2-x^2+n)/(2x)
(x^2+n)/(2x), where you can choose any starting value of x that you desire (though convergence to an exact (if possible) solution will be swifter the closer xi is to the actual value x)
In this case the number, n=95.54, so a decent starting value for x would be 10.
Using this initial x in (x^2+95.54)/(2x) will result in the following iterative sequence of x.
10, 9.777, 9.774457, 9.7744565, 9.7744565066299210578124802523397
The calculator result for my calc is: 9.7744565066299210578124802523381
So you see how accurate the newton method is in just a few iterations. :P
