Please I need help is for today
2 answers:
You might be interested in
Solve for <em>x</em> when √(<em>x</em> ² - 4) = 1 :
√(<em>x</em> ² - 4) = 1
<em>x</em> ² - 4 = 1
<em>x</em> ² = 5
<em>x</em> = ±√5
We're looking at <em>x </em>≤ 0, so we take the negative square root, <em>x</em> = -√5.
This means <em>f</em> (-√5) = 1, or in terms of the inverse of <em>f</em>, we have <em>f</em> ⁻¹(1) = -√5.
Now apply the inverse function theorem:
If <em>f(a)</em> = <em>b</em>, then (<em>f</em> ⁻¹)'(<em>b</em>) = 1 / <em>f '(a)</em>.
We have
<em>f(x)</em> = √(<em>x</em> ² - 4) → <em>f '(x)</em> = <em>x</em> / √(<em>x</em> ² - 4)
So if <em>a</em> = -√5 and <em>b</em> = 1, we get
(<em>f</em> ⁻¹)'(1) = 1 / <em>f '</em> (-√5)
(<em>f</em> ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5
The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of <em>f</em> at <em>x</em> = 1.
9514 1404 393
Answer:
possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}
actual roots: -1, (2 ±4i√2)/3
no turning points; no local extrema
end behavior is same-sign as x-value end-behavior
Step-by-step explanation:
The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.
The Rational Root Theorem tells us any rational roots will be of the form ...
±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}
= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots
Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.
The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.
The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):
f(x) = (x +1)(3x^2 -4x +12)
The complex roots of the quadratic can be found using the quadratic formula:
x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6
x = (2 ± 4i√2)/3 . . . . complex roots
__
The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.
