Tangents and smooth fit for x ln x − 3x/2

Text

Let $f(x)=x\ln x-\tfrac32 x$, with $x>0$. (a) Compute $f'(x)$ and determine any relative maxima and minima. (b) Show that there is exactly one tangent line to the graph drawn from the point $Q(0,-1)$: find its equation and the point of tangency. (c) Find $h,k$ such that $y=f(x)$ and the line $y=x+hx+k$ are tangent at the common point with abscissa $x=1$.

Solution

(a) Derivative: $f'(x)=\ln x+x\cdot\frac1x-\frac32=\ln x+1-\frac32=\ln x-\frac12.$ $f'(x)=0\Rightarrow \ln x=\tfrac12\Rightarrow x=\sqrt e\approx 1.6487$. Since $f''(x)=\dfrac1x>0$, it is a minimum: $f(\sqrt e)=\sqrt e\left(\tfrac12-\tfrac32\right)=-\sqrt e\approx -1.6487.$

(b) The tangent at the point $(t,f(t))$ has equation $y=f(t)+f'(t)(x-t)$. Requiring it to pass through $Q(0,-1)$: $-1=f(t)-t\,f'(t)=\big(t\ln t-\tfrac32 t\big)-t\big(\ln t-\tfrac12\big)=-t.$ Hence $t=1$ (the only value): the point of tangency is $(1,f(1))=(1,-\tfrac32)$. With $f'(1)=\ln1-\tfrac12=-\tfrac12$, the line is $y=-\tfrac12(x-1)-\tfrac32=-\tfrac12 x-1.$

(c) The line is $y=(1+h)x+k$, with slope $1+h$. Tangency at $x=1$ requires equal slopes and equal values: $1+h=f'(1)=-\tfrac12\ \Rightarrow\ h=-\tfrac32;\qquad (1+h)\cdot1+k=f(1)=-\tfrac32.$ From the second: $-\tfrac12+k=-\tfrac32\Rightarrow k=-1$. $\boxed{\;x_{min}=\sqrt e,\ \ \text{tangent from }Q:\ y=-\tfrac12x-1\ \text{at }(1,-\tfrac32),\ \ h=-\tfrac32,\ k=-1\;}$