The derivative of the function aa is given by a′(t)=2+9e0.4sinta′(t)=2+9e0.4sin⁡t , and a(1.2)=7.5a(1.2)=7.5. if the linear approximation to a(t)a(t) at t=1.2t=1.2 is used to estimate a(t)a(t), at what value of tt does the linear approximation estimate that a(t)=15a(t)=15 ?

  t ≈ 1.697798

Step-by-step explanation:

The linear approximation at t=1.2 is ...

  A ≈ A'(1.2)(t -1.2) +A(1.2)

  ≈ 15.0663(t -1.2) +7.5

We want this to be 15, so we have ...

  15 = 15.0663(t -1.2) +7.5

  7.5/15.0663 = t -1.2 . . . . subtract 7.5, divide by 15.0663

  7.5/15.0663 +1.2 = t ≈ 1.697798

The linear approximation at t = 1.2 is 1.697798.

The linear approximation at t = 1.2 is ...

A ≈ A'(1.2)(t -1.2) +A(1.2)

≈ 15.0663(t -1.2) +7.5

We want this to be 15, so we have ...

15 = 15.0663(t -1.2) +7.5

7.5/15.0663 = t -1.2                 subtract 7.5, divide by 15.0663

7.5/15.0663 +1.2 = t ≈ 1.697798

Therefore, The linear approximation at t = 1.2 is 1.697798.

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