
Find a function of the form f(x)=a+b√x .

Answer to Problem 9PS
f(x)=1+3√x
Explanation of Solution
Given information:
Find a function of the form f(x)=a+b√x that is tangent to the line 2y−3x=5 at the point (1,4) .
Calculation:
Consider the following function and its tangent at the point (1,4)
f(x)=a+b√x ,
2y−3x=5
To determine the values of the constants in the function, proceed as follows.
Differentiate the function.
f'(x)=limh→0f(x+h)−f(x)h=limh→0(a+b√x+h)−(a+b√x)h
=limh→0b(√x+h−√x)h
Rationalize with (√x+h+√x)
f'(x)=limh→0b(√x+h−√x)h×[(√x+h+√x)(√x+h+√x)]
=limh→0b((x+h)−x)h(√x+h+√x)=limh→0bhh(√x+h+√x)
=b√x+√x
Hence, f'(x)=b2√x
Now find the value of the derivative at the point (1,4)
f'(x)=b2√x
m=f'(1)=b2
Now find the equation of the tangent at the point (1,4)
y−y1=m(x−x1)y−4=b2(x−1)y=bx2−b2+4
Now rearrange the equation in the slope intercept form.
Slope-intercept form of a line y=mx+c
2y−3x=52y=3x+5
y=32x+52
Now compare the equations.
b2=32b=3
As the point (1,4) lies on the graph of the function
f(x)=a+b√xf(1)=a+b√14=a+b
Now substitute the value b and solve for a
4=a+3a=1
Hence, the expression for the function is f(x)=1+3√x .
Chapter 12 Solutions
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