$F(x) = 2x + 4^{-x}$ . Show that the tangent to the curve with $y = F(x)$ at the point at $(-1,y)$ is $15(\ln 2) x + 2y + 15(\ln 2) - 9 = 0$

Question

I am struggling to manipulate the following into the required solution

Answer 1

The slope of the tangent to a curve can be found by differentiating the function: \begin{align} F(x)&=2x+4^{-x}\\ F'(x)&=2-\ln4\cdot4^{-x}\\ F'(-1)&=2-\ln4\cdot4\\ F'(-1)&=2-8\ln2 \end{align}

Once we have the slope, we can find the point: \begin{align} y=F(-1)&=-2+4\\ y&=2 \Rightarrow (-1,2) \end{align}

Substitute the point and the slope into the equation of a line: \begin{align} y-y_1&=m(x-x_1)\\ y-2&=(2-8\ln2)(x+1)\\ y-2&=(2-8\ln2)x+2-8\ln2\\ 0&=(2-8\ln2)x-y+4-8\ln2 \end{align}

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The given function should instead be $$F(x)=2^x+4^{-x}.$$

Below, the same steps with a different function. \begin{align} F(x)&=2^x+4^{-x}\\ F(x)&=2^x+2^{-2x}\\ F'(x)&=\ln2\cdot2^x-2\ln2\cdot2^{-2x}\\ F'(-1)&=\ln2\cdot\frac12-2\ln2\cdot2^2\\ F'(-1)&=-\frac{15}2\ln2 \end{align}

\begin{align} y=F(-1)&=\frac12+4\\ y&=\frac92\Rightarrow(-1,\frac92) \end{align}

\begin{align} y-y_1&=m(x-x_1)\\ y-\frac92&=-\frac{15}2\ln2\cdot(x+1)\\ 2y-9&=-15\ln2\cdot(x+1)\\(15\ln2)x+2y-9+15\ln2&=0 \end{align}

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