help! solve one problem (1 Viewer)

[zhouyuezy]

Hi, everyone, is there anyone can help me with the following question? Thank you!
let f(x)=3^x - x^3. The tangent to the curve is parallel to the secant through (0,1) and (3,0) for x equal
(A) only to 0.984
(B) only to 1.244
(C) only to 2.727
(D) to 0.984 and 2.804
(E) to 1.244 and 2.727

I need the procedure and result.
Thank you all~~
Happy Xmas!

 

[Affinity]

Looks like an assessment question, it has something to do with the derivative, but you would have to resort to numerical methods such as halving the interval. This might help:

 

[jyu]

Hi, everyone, is there anyone can help me with the following question? Thank you!
let f(x)=3^x - x^3. The tangent to the curve is parallel to the secant through (0,1) and (3,0) for x equal
(A) only to 0.984
(B) only to 1.244
(C) only to 2.727
(D) to 0.984 and 2.804
(E) to 1.244 and 2.727

I need the procedure and result.
Thank you all~~
Happy Xmas!

f(x)=3^x - x^3
f(x) = e^(xln3) - x^3
f'(x) = (ln3)e^(xln3) - 3x^2 = -1/3
(ln3)e^(xln3) - 3x^2 + 1/3 = 0
Graph y = (ln3)e^(xln3) - 3x^2 + 1/3
Find x-intercepts 1.244 and 2.727

 

[zhouyuezy]

Is that the graph is the only way to solve?
Thank you all~

 

[Affinity]

There are others.