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:

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....
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Answer 2

This is what I got.

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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