Answer:
b=2
Step-by-step explanation:
b-2=0
b=2
![\bf f(x)=(x-6)e^{-3x}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)] \\\\\\ \cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D%28x-6%29e%5E%7B-3x%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%5Ccdot%20e%5E%7B-3x%7D%2B%28x-6%29-3e%5E%7B-3x%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3De%5E%7B-3x%7D%5B1-3%28x-6%29%5D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3De%5E%7B-3x%7D%2819-3x%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B19-3x%7D%7Be%5E%7B3x%7D%7D)
set the derivative to 0, solve for "x" to get any critical points
keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there
there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema
Answer:
a. 7
b. 5
c. 0
d. 2
Step-by-step explanation:
just place the same number in each box
Number 1 is neither
Number 2 is B
Hi there!
The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation.
x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}
x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}
x = \frac{-5 \pm \sqrt{25 - 12} }{2}
x = \frac{-5 \pm \sqrt{13} }{2}
x = \frac{-5 \pm 3.61 }{2}
x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}
x=-0.695 \ \textgreater \ \ \textgreater \ -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31
Therefore, the solutions to the quadratic equation are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!
Your answer is 4.31
