If F(x)=x^3−7x+5, use the limit definition of the derivative to find F�Œ(5), then find an equation of the tangent line to the curve y=x^3−7x+5 at the point (5, 95).
F�Œ(5)=
The equation of the tangent line is y = x + .
Check your answer for yourself by graphing the curve and the tangent line.
f(5) = 125 - 35 + 5 = 95
f(5+h) = (5+h)^3 - 7(5+h) + 5
= h^3 + 15h^2 +68h + 95 --- (I'll let you check that
derivative = Lim ( f(h+5) - f(5) )/h as h ---> 0
= lim ( h^3 + 15 h^2 + 68h = 95 - 95)/h
= lim h^2 + 15h + 68 , as h --> 0
= 68
so equation of tangent at (5,95)
y - 95 = 68(x-5)
y = 68x - 340 + 95
y = 68x - 245
To find F'(x) using the limit definition of the derivative, we need to apply the formula:
F'(x) = lim(h->0) [F(x+h) - F(x)] / h
First, let's find F�Œ(5). Plug in x=5 into the function F(x):
F(5) = (5)^3 - 7(5) + 5
= 125 - 35 + 5
= 95
Next, let's find F(x+h):
F(x+h) = (x+h)^3 - 7(x+h) + 5
= x^3 + 3x^2h + 3xh^2 + h^3 - 7x - 7h + 5
Now, we can calculate F�Œ(5) using the limit definition of the derivative:
F�Œ(5) = lim(h->0) [F(x+h) - F(x)] / h
= lim(h->0) [(x^3 + 3x^2h + 3xh^2 + h^3 - 7x - 7h + 5) - (x^3 - 7x + 5)] / h
= lim(h->0) [3x^2h + 3xh^2 + h^3 - 7h] / h
= lim(h->0) 3x^2 + 3xh + h^2 - 7
= 3x^2 - 7
Now, plug in x=5 into the equation F�Œ(x):
F�Œ(5) = 3(5)^2 - 7
= 75 - 7
= 68
Therefore, F�Œ(5) = 68.
To find the equation of the tangent line to the curve at the point (5, 95), we can use the point-slope form of a line:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope of the tangent line.
The slope of the tangent line is equal to F�Œ(5), which we found to be 68. Substituting the values into the equation, we have:
y - 95 = 68(x - 5)
Now, simplify and rearrange the equation to get it in slope-intercept form:
y - 95 = 68x - 340
y = 68x - 245
Therefore, the equation of the tangent line is y = 68x - 245.
To check your answer, you can graph the curve y = x^3 - 7x + 5 and the tangent line y = 68x - 245. If they intersect at the point (5, 95), you have the correct equation for the tangent line.