Question

When x is real, the greatest possible value of ${10}^x-{100}^x$ is

Answer 1

Given, $f\left(x\right)={10}^x-{100}^x$

This can be rewritten as

$f\left(x\right)={10}^x-{10}^{2x}$

On differentiating the $f\left(x\right)$, we get

$f^{\prime }\left(x\right)={10}^x\cdot \log 10-{10}^{2x}\cdot \log 100$

$f^{\prime }\left(x\right)=\log 10\cdot ({10}^x-{10}^{2x})$

for extremum, $f^{\prime }\left(x\right)=0$, hence

${10}^x=2\cdot {10}^{2x}$

$\Rightarrow 1=2\cdot {10}^x$

taking log on each side,

$\Rightarrow x=\log 1/2=-0.301$

Now, putting the value of this $x$ into the $f\left(x\right)$ we get $f\left(x\right)=0.25$

For confirmation, we can check its the point of maximum by finding and showing $f^{\prime \prime }\left(x\right)<0$ at $x=-0.301$, which means that its point of maxima.

Hence, we can say that the Maximum value of the given function is $0.25$