For f(x)=\sqrt{x}^{3}, what is the equation of the line tangent

Siena Duran

Siena Duran

Answered question

2022-02-10

For f(x)=x3, what is the equation of the line tangent to the graph of f at the point where x=125?

Answer & Explanation

Javon Ross

Javon Ross

Beginner2022-02-11Added 13 answers

Explanation:
At x=125,y=f(125)=(125)3=(55)3=12555=6255
f(x)=x32, so f(x)=32x12 and
at x=125, the slope of the tangent line is m=f(125)=1552
The equation of the line through the point (125,6255) with slope m=1552 is
y=1552x62552. (In slope-intercept form).
czapkowyrqo

czapkowyrqo

Beginner2022-02-12Added 7 answers

Explanation:
f(x)=(x)3
f(x)=x32
To find the equation of the tangent line, we need a point and a slope. The point would be at (125, f(125)), and the slope is f'(125)
Point: (125, f(125))
(125,12532)
(125, 1397.54)
Slope:
f(x)=32x12
f(125)=32125
=1525
Equation:
y12532=1525(x125)

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