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I want to find the equation of the tangent to $f(x)$ that it's slope is the greatest of all.

$$f(x) = 2e^x - \frac14e^{2x}$$

To find the highest slope I do $f''(x) = 0$ ($f'(x)$ is the function that gives the slope so I need to do again derivative).

$$f''(x) = e^x(2-e^x)$$ $$x = \ln2$$

However, in the answers, it's $2x-2ln2+3$, i.e. the slope is $2$.

What am I doing wrong?

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    So what is the slope at x = ln 2? $f'(x) = 2e^x - \frac 14e^{2x}2$ so $f'(\ln 2) = 2*2 - \frac 14 2^2*2 = 2$ so the slope is 2. – 2017-02-18

2 Answers 2

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You are right that the minimum is reached at $x_0=\ln(2)$. Then, the slope is given by $f'(x_0)$, which is $2e^{\ln(2)}-0.25e^{2\ln(2)}=4-0.25e^{\ln(4)}= 2$.

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    So basically what I found it's the tangent point, not the slope. – 2017-02-19
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    Yes, or more precisely: you found the $x$-coordinate of the point where the slope is maximal. – 2017-02-19
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$f(x) = 2e^x - \frac14e^{2x}$

$f'(x) = 2e^x - \frac 14e^{2x}*2 = 2e^x - \frac 12e^2x$

$f''(x) = 2e^x - e^2x; f''(x) = 0 \iff x = \ln 2$

$f'''(x) = 2e^x - 2e^{2x}; x = \ln 2 \implies f'''(x) = 2*2-2(2^2) < 0$

$f''(\ln 2) = 0 \implies x=\ln 2; f'(\ln 2) = 2*2 - \frac 12 2^2 = 2; f(\ln 2) = 2*2 -\frac 14 2^2 = 3$ is an extrema for $f'(x)$; the slope of $f(x)$.

$f'''(\ln 2) < 0 \implies x=\ln 2; f'(\ln 2) = 2$ is a maximum value for $f'(x)$; the slope of $f(x)$.

So the sharpest tangent line is at $x_0 = \ln 2$ where the slope is $f'(\ln 2)=2$ so the equation of the line is:

$y- f(x_0) = f'(x_0)(x - x_0)$

$y - 3 = 2(x-\ln 2)$

$y = 2x - 2\ln2 + 3$

So what did you do wrong? You confused "$x = \ln 2$ so slope $= f'(x)= 2$" with "$x = \ln 2$ so slope $ = x = 2$", apparently.