If f(x)=2x^4-x^3-3x^2+14 then f'(3)=
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Answer:
a) The coordinates are (0.431, -2.646) and (3.568,-7.354)
b) The tangent lines are
l₁ = (x-0.431)*2/3 - 2.646
l₂ = (x-3.568)2/3 - 7.354
Step-by-step explanation:
First, lets complete squares, by taking for each cordinate the square of a linear expression
0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8
Hence (x-2)² + (y+5)² = 8
Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.
(y+5)² = 8 - (x-2)² = -x² + 4x + 4
Thus
Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.
We may just find the values of x which the derivate of f is either 2/3 or -2/3.
The quadratic has roots
one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.
Also g(3.568) = - 7.354 and f(0.431) = -2.646
This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are
l₁ = (x-0.431)*2/3 - 2.646
l₂ = (x-3.568)2/3 - 7.354