Our function. y=(x^2 ) +6 The inverse function switches from (x,y) to (y, x) so where you see y put a x and where you see x put a y. Inverse function. X=(y^2)+6 Now we still needed to have it in the form of y= something so lets work it. X-6=y^2 Square root (x-6)=y f^-1 (x)=square root (x-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.
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<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>
For every integer k from 1 to 10, inclusive the kth term of a certain sequence is given by [(−1)^(k+1)]∗[1/2^(k)]. If T is the sum of the first 10 terms in the sequence, then T is:
B.
A. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4
