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Sagot :
In order to find the line tangent to f(x) and parallel to the given equation, first let's find the slope of the given line, putting it in the slope-intercept form:
[tex]y=mx+b[/tex]Where m is the slope and b is the y-intercept.
So we have:
[tex]\begin{gathered} 15x-3y=8 \\ -3y=8-15x \\ 3y=15x-8 \\ y=5x-\frac{8}{3} \end{gathered}[/tex]So the slope is m = 5.
Since parallel lines have the same slope, the slope of the tangent line is equal to 5.
In order to find when the slope of the tangent line is equal to 5, let's calculate the derivative of f(x). This derivative will give the slope in function of x:
[tex]\begin{gathered} m=f^{\prime}(x) \\ m=3^x\cdot\ln 3 \\ m=1.1\cdot3^x \end{gathered}[/tex]Equating the slope to 5, we have:
[tex]\begin{gathered} 1.1\cdot3^x=5^{} \\ 3^x=4.545^{} \\ \ln (3^x)=\ln (4.545) \\ x\cdot\ln (3)=\ln (4.545) \\ x=\frac{\ln (4.545)}{\ln (3)}=\frac{1.5141}{1.099}=1.378 \end{gathered}[/tex]Calculating the value of y for x = 1.378, we have:
[tex]\begin{gathered} y=3^{1.378} \\ y=4.544 \end{gathered}[/tex]So the tangent line passes through the point (1.378, 4.544).
Using this point, let's find the y-intercept of the tangent line:
[tex]\begin{gathered} y=mx+b \\ 4.544=5\cdot1.378+b \\ 4.544=6.89+b \\ b=4.544-6.89 \\ b=-2.346 \end{gathered}[/tex]So the tangent line is y = 5x - 2.346.
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