An equation of the line tangent to the graph of f(x) = (4x2-8x+3)4 at the point where x = 1is

Question

 A. x=1

B. y=1

C. y-81 384(x-1)

D. y-1=3.37497(x-1)

E. y=5(x-1)

thinking its B cant be sure though

Answer 1

Yes the equation of the tangent line is y=1.

The point on the graph of f where x=1 is the point (1, 1) because f(1) = (4·1² - 8·1 + 3)⁴ = (-1)⁴ = 1.

The derivative of f at this point is 0 because f'(x) = 4(4x² - 8x + 3)³(8x - 8), so f'(1) = 0

The equation of the straight line of slope 0 through the point (1, 1) (or indeed through any point whose y coördinate is 1) is y=1.

Answer 2

If we are talking about the equation f(x) = (4x²-8x+3)⁴ ...

f'(x) = 4(4x²-8x+3)³ ⋅(8x-8) by Chain Rule and when x = 1 the slope of the tangent line (f'(x)) is = 0 (horizontal line)

Then, we can plug-in x = 1 into original equation f(x) to find a point on that line:

So, f(1) = (4(1)²-8(1)+3)⁴ .= (-1)⁴ = 1

This means a horizontal tangent line at the ordered pair: (1,1) would give y = 1 or Answer B.

Hope this helps,

Todd