Question

Consider f(x)=sqrtx. Find all value(s), c , in the interval [0,1] such that the slope of the tangent line to the graph of f at c is parallel to the secant line through the points (0,f(0)) and (1,f(1)),

Answer 1

first we need the slope of the line between \((0,f(0))\) and \((1,f(1))\)
we have \(f(x)=\sqrt{x}\) so \(f(0)=\sqrt{0}=0\) and \(f(1)=\sqrt{1}=1\)
slope is therefore 1

Answer 2

now we need the formula for the slope of the tangent line
that is the derivative, and the derivative of \(f(x)=\sqrt{x}\) is \(f'(x)=\frac{1}{2\sqrt{x}}\)

Answer 3

finally we need the value of \(c\) for which the derivative is equal to the slope of the secant line, so we set
\[\frac{1}{2\sqrt{c}}=1\] and solve for \(c\)
that should take at most two steps

Answer 4

thanks a bunch!