What is the slope of the tangent line to the graph of the function f(x) = x^3 – 7x + 8 at the point (2, 2)?
2007-04-04 04:52:52 · 4 answers · asked by Laguna B 1 in Science & Mathematics ➔ Mathematics
The slope of the tangent line is the derivative. So just take the derivative and plug in x=2:
f(x) = x^3 - 7x + 8
f ' (x) = 3x^2 - 7
f ' (2) = 3(2)^2 - 7
f ' (2) = 12 - 7
f ' (2) = 5
Your answer is 5!
2007-04-04 04:59:01 · answer #1 · answered by johnny 2 · 0⤊ 0⤋
Since the slope at EVERY point on that line is the derivative of the function:
f(x) = x^3 – 7x + 8
f'(x) = 3x^2 – 7
Since we want the slope at only one point, (2,2):
f'(2) = 3(2)^2 - 7
f'(2) = 5
Therefore, the slope of the line tangent to the graph of the functions is 5.
2007-04-04 11:57:43 · answer #2 · answered by HallamFoe 4 · 0⤊ 0⤋
Differentiate
f'(x) = 3x^2 - 7
Then put in the x you want to find the slope at (x = 2)
find f'(2)
2007-04-04 11:55:25 · answer #3 · answered by Orinoco 7 · 0⤊ 0⤋
dy/dx = 3x^2 -7
at (2,2) it is 5.
2007-04-04 11:58:08 · answer #4 · answered by gebobs 6 · 0⤊ 0⤋