Can someone explain this question?

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2 Answers
Gió
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'(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....

A08
Mar 9, 2017

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)

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