first off, is noteworthy that's the graph of an exponential function, thus the function will be along the lines of g(x) = abˣ , now, what's "a" and "b" values?

well, let's take a peek when x = 0 and x = 1.

$\bf g(x) = ab^x \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 0\\ y = 1 \end{cases}\implies 1=ab^0\implies 1=a(1)\implies \boxed{1=a} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 1\\ y = 4 \end{cases}\implies 4 = ab^1\implies 4=1b^1\implies \boxed{4=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill g(x) = 4^x\qquad \qquad \qquad \begin{array}{|c|c|ll} \cline{1-2} x&y\\ \cline{1-2} -2&\frac{1}{4^2}\to \frac{1}{16}\\ -1&\frac{1}{4}\\ 0&1\\ 1&4\\ 2&16\\ \cline{1-2} \end{array}~\hfill$

8 0

3 years ago

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

3 years ago