Question

Can someone explain this question?

Answer 1

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'(t)=e^(2t^2)3*10^(3t-5)ln(10)+10^(3t-5)*4te^(2t^2)`
rearranging:
`f'(t)=e^(2t^2)*10^(3t-5)[3ln(10)+4t]`
Let us set this equal to zero:
`e^(2t^2)*10^(3t-5)[3ln(10)+4t]=0`
when:
`t=-(3ln(10))/4` that makes the square bracket equal to zero....
I think....

Answer 2

This is what I got.

Explanation:

Given expression is
`f(t)=10^(3t-5)xxe^(2t^2)`
Using product rule
`f'(t)=10^(3t-5)xx(e^(2t^2)xx4t)+e^(2t^2)xxln10xx10^(3t-5)xx3`
`=>f'(t)=(4t+3ln10)xx10^(3t-5)xxe^(2t^2)`

To meet the given condition
`(4t+3ln10)xx10^(3t-5)xxe^(2t^2)=0` .....(1)
`=>f(t)[4t+3ln10]=0`
either `f(t)=0`
or`[4t+3ln10]=0`
`=>t=-(3ln10)/4` ....(2)

To find roots of `f(t)=0`,
`10^(3t-5)xxe^(2t^2)=0`
either `10^(3t-5)=0` .......(3)
or `e^(2t^2)=0` ......(4)

From both (3) and (4)
we get `t=-oo`.
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)