# Can someone explain this question?

Mar 9, 2017

You need to derive your function with respect to $t$ and then solve the resulting equation when you set the derivative equal to zero:

#### Explanation:

Ok, I can be wrong but your function can be derived as:
$f ' \left(t\right) = {e}^{2 {t}^{2}} 3 \cdot {10}^{3 t - 5} \ln \left(10\right) + {10}^{3 t - 5} \cdot 4 t {e}^{2 {t}^{2}}$
rearranging:
$f ' \left(t\right) = {e}^{2 {t}^{2}} \cdot {10}^{3 t - 5} \left[3 \ln \left(10\right) + 4 t\right]$
Let us set this equal to zero:
${e}^{2 {t}^{2}} \cdot {10}^{3 t - 5} \left[3 \ln \left(10\right) + 4 t\right] = 0$
when:
$t = - \frac{3 \ln \left(10\right)}{4}$ that makes the square bracket equal to zero....
I think....

Mar 9, 2017

This is what I got.

#### Explanation:

Given expression is
$f \left(t\right) = {10}^{3 t - 5} \times {e}^{2 {t}^{2}}$
Using product rule
$f ' \left(t\right) = {10}^{3 t - 5} \times \left({e}^{2 {t}^{2}} \times 4 t\right) + {e}^{2 {t}^{2}} \times \ln 10 \times {10}^{3 t - 5} \times 3$
$\implies f ' \left(t\right) = \left(4 t + 3 \ln 10\right) \times {10}^{3 t - 5} \times {e}^{2 {t}^{2}}$

To meet the given condition
$\left(4 t + 3 \ln 10\right) \times {10}^{3 t - 5} \times {e}^{2 {t}^{2}} = 0$ .....(1)
$\implies f \left(t\right) \left[4 t + 3 \ln 10\right] = 0$
either $f \left(t\right) = 0$
or$\left[4 t + 3 \ln 10\right] = 0$
$\implies t = - \frac{3 \ln 10}{4}$ ....(2)

To find roots of $f \left(t\right) = 0$,
${10}^{3 t - 5} \times {e}^{2 {t}^{2}} = 0$
either ${10}^{3 t - 5} = 0$ .......(3)
or ${e}^{2 {t}^{2}} = 0$ ......(4)

From both (3) and (4)
we get $t = - \infty$.
This solution is purely theoretical as such this equation has no solution in real or complex numbers.

As such only possible solution is as given in equation (2)